Noindent 240 <Quad J . Vukman <[< Begin 1 Aligned L Cfwl

240 .. J period Vukman \noindentLine 1 C-F-w-L240 \ o-h-o-e-rquad J t-m-i . Vukman sub c-m y-h-b h-g-n-b to the power of e-i-i sub n-v-e y-e-g-a s-brackleft b-o-one-parenleft sub three-x-v-e c-e-parenright-parenright comma-a r-w-u-x-e l-i-brackright-s t-e-a i-o-t-n o-parenleft-f-parenleft parenleft-one-n to the power\ [ \ begin of four-one-six{ a l i g n e d parenright-one-a} C−F−w−L on-period-parenright-w−h−o−e−r t−m−i Line{ 2c open−m } parenthesisy−h−b 1 7 h closing−g−n−b parenthesis ˆ{ e−i− bracklefti { n− Tv− parenleft-xe y−e−g−a closings−b r a parenthesisc k l e f t }} commab−o− xone brackright−p a r e n x l e y f brackleftt { three parenleft-T−x−v−e x closing c−e− parenthesisparenright comma−parenright x brackright x comma = 0 comma−a x r− commaw−u−x−e yl− subi−brackright in R sub period−s t−e−a } i−o−t−n o−p a r e n l e f t −f−parenleft parenleft −one−n ˆ{ four−one−s i x parenrightRi g-h t m− utone p− la c-i} a t-in− onperiod .. f open−parenright parenthesis− 1w closing\\ parenthesis b y brackleft T open parenthesis x comma-parenright x (open 1 parenthesis 7 ) 1 8 closing [ T parenthesis parenleft brackleft-brackleft−x ) , T open x parenthesis ] x x comma y [ x comma parenleft x x brackleft−T T x open ) parenthesis , x x] closing x parenthesis = 0 comma , x sub ,x brackright y {\ Rowin } 1 commaR { Row. }\ 2end 0 .{ ina lR i g period n e d }\ ] 240 J . Vukman Sub t-r ac i-t ng open parenthesis 18 closing parenthesis fro m open parenthesis 7 closing parenthesis o neo b-t a n s brackleft brackleft T open parenthesis x comma-parenright x comma x brackright y \ centerline {Ri $ g−h $ t m ut p l $ c−i $ a $ t−i $ on \quad f(1)by$[ T ( x e−i−in−v−ey−e−g−as−brackleft four−one−sixparenright−one−a a to theC power− F of− w n ..− dLo ..− rh i− too the− e power− rt − ofm m− eic ..−m ny ..− Rh ..− hbh ..− pg − n − b b − o − one − parenleftthree−x−v−ec−e−parenright−parenrightcomma−ar−w−u−x−el−i−brackright−st−e−ai − o − t − no − parenleft − f − parenleftparenleft − one − n n − period − parenright − w commaopen−parenright parenthesis x open x parenthesis $ } (17) [T parenleft − x), x]xy[parenleft − T x), x]x = 0, x, y R T .. h e .. open parenthesis 9 to the power of closing parenthesis g .. v ∈ . \ [( 1 8 ) brackleft −brackleft T ( x , x , x x [ T ( x ) , { x } Line 1 brackleft to theRi g power− h t of m brackleft ut p l c − subi a Tt − openi on parenthesis f ( 1 ) b ysub [T ( closingxcomma parenthesis− parenrightx to the power of x comma brackright y] brackright =\begin plus{ array brackleft}{ c} brackleft, \\ 0 sub\end T x{ toarray the} powery \ in of closingR. parenthesis\ ] y brackright comma x plus brackleft brackleft sub T open parenthesis Line 2 open parenthesis 2 0 closing parenthesis comma y, brackright plus brackleft sub brackleft T open (18) brackleft − brackleftT (x, x, xx[T (x), ] = y ∈ R. parenthesis sub x closing parenthesis comma y to the power of brackright commax x0 brackright plus sub brackleft brackleft T open parenthesis\noindent toSub the power $ t− ofr y $ comma ac sub $ i x− brackrightt$ ng(18) = 0 comma from(7)oneo $b−t $ a n s $ [ [ T Sub t − r ac i − t ng ( 18 ) fro m ( 7 ) o neo b − t a n s [[T (xcomma − parenrightx, x]y an d r (P-w x sub comma i h-t-t− tparenright sub ngh e sub ab x o to the , power x of ] x v-f y$ to the power of e-o r r l sub itn 0 to the power of w-parenright e a riv im e n R h p i a-n$ sub a ˆ g-t{ n open}$ parenthesis\quad d 1\ closingquad parenthesisr $ i ˆ{ commam }$ open e parenthesis\quad n \ toquad the power$ R of $ parenright-nine\quad h \quad a n d openp parenthesis 2 0 closing parenthesis w eo .. b sub t a i sub n ( x ( \ [Line ( 1 x plus brackleft ( \ ] T open parenthesis x closing parenthesis comma x y x plus open parenthesis closing parenthesis y brackright Line 2 plus brackleft brackleft xT open parenthesisT y closing h e parenthesis (9) g v comma sub x brackright x-comma x brackright = to the power of x brackleft sub brackleft \ centerline {T \quad h e \quad $ ( 9 ˆ{ ) }$ g \quad) v } y] + [[T x y], x + [[T ( ] y \ [ \ begin { a l i g n e d } [ ˆ{ [ } { (20)T } , y(] + ˆ{ [x[T (}x){, y ), x]} +,[ [T ( ],x ] y = 0, ] + [ [ { T } x ˆ{ ) } y ] , x + [ { [ } { T(x }\\e−or w−parenrighte parenright−nine P − wih − t − ttngheabov − f r litn0 a riv i a − ng−t(1), ( a n d ( 2( 2 0 0 ) w ) eo ,b a yi ] + [ { [ } T( { x } ) , y ˆ{ ] } , x ] + { [ } [ T ( ˆ{ y } , t { nx } ] = 0 , \end{ a l i g n e d }\ ] +[T (x), xyx + ()y]

x[[ \ hspace ∗{\ f i l l } $ P−w { i } h−t−+[[txT ( ty),x{]xngh− commax} e] ˆ{ =x } { ab o } v−f ˆ{ e−o r }$ r $ l { i t n } 0 ˆ{ w−parenright e }$ a r i v i $ a−n { g−t } ({ 1 } ) , ( ˆ{ parenright −nine }$ and(20)weo \quad $ b { t }$ a $ i { n }$

\ [ \ begin { a l i g n e d } +[T(x),xyx+()y] \\ + [ [ { xT } ( y ){ , } { x } ] x−comma x ] = ˆ{x{ [ } { [ }}\end{ a l i g n e d }\ ] Centralizers on semiprime rings .. 241 CentralizersWe have therefore on semiprimebrackleft T open rings parenthesis\quad x241 closing parenthesis comma x brackright yx to the power of 2 minus x to the powerWehavetherefore of 2 y brackleft T open $[ parenthesis T ( x closing x parenthesis ) , commax ] x brackright yxˆ{ 2 plus} xyx − brackleftx ˆ{ 2 T} openy parenthesis [ T x closing ( x parenthesis) ,x]+xyx[T(x) comma x brackright minus brackleft T open parenthesis ,x] x closing parenthesis− [T(x) comma x brackright xyx ,x = 0 comma ] xyx =x 0comma , y$ in R comma which reduces because of open parenthesis 9 closing parenthesis and open parenthesis 1 5 closing parenthesis to \noindentbrackleft T open$ x parenthesis , y x closing\ in parenthesisR , $ comma which x brackright reduces yx because to the power of ( of 9 2 minus ) and x to ( the 1 5 power ) to of 2 y brackleft T open parenthesis x closing parenthesis comma x brackright = 0 comma x comma y in R period \ [[T(x)Left multiplication of the above ,x relation ] by x yxˆ gives{ 2 } − x ˆ{ 2 } y [ T ( x ) , x ] Centralizers on semiprime rings 241 We have therefore [T (x), x]yx2 − x2y[T (x), x] + xyx[T (x), x] − =x brackleft0 , T x open , parenthesis y \ in x closingR. parenthesis\ ] comma x brackright yx to the power of 2 minus x to the power of 3 y [T (x), x]xyx = 0, brackleft T open parenthesis x closing parenthesis comma x brackright = 0 comma x comma y in R period x, y ∈ R, which reduces because of ( 9 ) and ( 1 5 ) to One .. can .. replace .. in .. the .. above .. relation comma .. according .. to .. open parenthesis 1 5 closing parenthesis comma x\noindent brackleft T openLeft parenthesis multiplication x closing parenthesis of the above comma relation x brackright by yx .. $ by x $ gives [T (x), x]yx2 − x2y[T (x), x] = 0, x, y ∈ R. xy brackleft T open parenthesis x closing parenthesis comma x brackright x comma which gives Equation: open parenthesis 2 1\ [x closing parenthesis [T(x)Left multiplication .. xy brackleft of the T ,xabove open relation parenthesis ] by yxˆx xgives closing{ 2 } parenthesis − x ˆ{ comma3 } xy brackright [ T x to the ( power x of ) 2 minus , x to the] power = of 0 3 y ,brackleft x T , open y parenthesis\ in xR. closing\ parenthesis] comma x brackright = 0 comma x comma y in R period Left multiplication of the above relationx[T (x by), xT]yx open2 − parenthesisx3y[T (x), x] x = closing 0, x, parenthesis y ∈ R. gives Equation: open parenthesis 22 closing parenthesis .. T open parenthesis x closing parenthesis xy brackleft T open parenthesis One can replace in the above relation , according to (15), x[T (x), x]yx by x\noindent closing parenthesisOne \quad commacan x brackright\quad r x e top l athe c e power\quad of 2in minus\quad T openthe parenthesis\quad above x closing\quad parenthesisr e l a tx i o to n the , power\quad ofaccording 3 y \quad to \quad brackleft$(15) T open parenthesis ,x[T(x) x closing parenthesis comma x brackright ,x]yx$ = 0 comma x comma\quad y inby R period The substitution T open parenthesis x closing parenthesisxy[ yT for(x), y x] inx, open whichgives parenthesis 2 1 closing parenthesis leads to \ beginEquation:{ a l i g open n ∗} parenthesis 23 closing parenthesis .. xT open parenthesis x closing parenthesis y brackleft T open parenthesis 2 3 xxy closing [ parenthesis T ( comma x x ) brackright ,xy x x[T to(x the) ], x]x power x− x ofy ,[ 2T ( minusx which), x] =x to 0, the gives x, power y ∈ R.\\\ of 3tag T open∗{$ parenthesis ( 2 1 x(21) closing ) $} parenthesisxy [ y Tbrackleft ( T x open ) parenthesis , x x closing ] x parenthesis ˆ{ 2 } comma − x x ˆbrackright{ 3 } y[T(x),x]=0 = 0 comma x comma y in R period Left multiplication of the above relation by T (x) gives ,Subtracting x , open y parenthesis\ in R. 23 closing parenthesis from open parenthesis 22 closing parenthesis we obtain \endbrackleft{ a l i g n T∗} open parenthesis x closing parenthesis comma x brackright y brackleft T open parenthesis x closing parenthesis comma x brackright x to the powerT (x of)xy 2[ minusT (x), x brackleft]x2 − T (x T)x3 openy[T (x parenthesis), x] = 0, x,x closingy ∈ R. parenthesis comma(22) x to the power of 3 brackright\noindent y brackleftLeft multiplication T open parenthesis of x closing the above parenthesis relation comma byx brackright $T = ( 0 comma x ) x comma$ gives y in R period From theThe above substitution relation andT (x Lemma)y for y 1in it ( follows 2 1 ) leads that to \ beginopen{ parenthesisa l i g n ∗} brackleft T open parenthesis x closing parenthesis comma x to the power of 3 brackright minus brackleft T open\ tag parenthesis∗{$ ( 22 x closing ) $ parenthesis} T(x)xy[T(x),x]xˆ comma x brackright x to the power of 2 closing parenthesis y brackleft{ 2 T} open − parenthesisT( x closing ) parenthesis x ˆ{ 3 } commay[T(x),x]=0,x,y x brackrightxT (x)y =[T ( 0x comma), x]x2 − xx comma3T (x)y[ yT ( inx) R, x comma] = 0, x, y ∈ R. \(23)in R. \endwhich{ a l i reduces g n ∗} to open parenthesisSubtracting x brackleft ( 23 ) from T open ( 22 parenthesis ) we obtain x closing parenthesis comma x brackright x plus x to the power of 2 brackleft T open\noindent parenthesisThesubstitution x closing parenthesis $Tcomma x ( brackright x ) closing y$ parenthesis for $y$ y brackleft in T (21) open parenthesis leads xto closing parenthesis comma x brackright = 0 comma[ xT ( commax), x]y[T y( inx) R, x] periodx2 − [T (x), x3]y[T (x), x] = 0, x, y ∈ R. \ begin { a l i g n ∗} RelationFrom .. open the above parenthesis relation 9 closingand Lemma parenthesis 1 it follows .. makes that it possible to write .. brackleft T open parenthesis x closing parenthesis\ tag ∗{$ ( comma 23 x brackright ) $} xT(x)y[T(x),x]xˆ x instead of x brackleft { 2 } − x ˆ{ 3 } T(x)y[T(x),x]=0,x,y \ in R. 3 2 \end{ a l i g n ∗} ([T (x), x ] − [T (x), x]x )y[T (x), x] = 0, x, y ∈ R, which reduces to \noindent Subtracting ( 23 ) from ( 22 ) we obtain (x[T (x), x]x + x2[T (x), x])y[T (x), x] = 0, x, y ∈ R. \ [[T(x),x]y[T(x),x]xˆ{ 2 } − [T( x ) ,Relation x ˆ{ (3 9} )]y[T(x),x]=0,x,y makes it possible to write [T (x), x]x instead of x[ \ in R . \ ]

\noindent From the above relation and Lemma 1 it follows that

\ [ ( [ T ( x ) , x ˆ{ 3 } ] − [ T ( x ) , x ] x ˆ{ 2 } ) y [T(x),x]=0,x,y \ in R, \ ]

\noindent which reduces to

\ [(x[T(x),x]x+xˆ{ 2 } [ T ( x ) , x ] ) y[T(x),x]=0,x,y \ in R. \ ]

\noindent Relation \quad ( 9 ) \quad makes it possible to write \quad $ [ T ( x ) , x ] x$ insteadof $x [$ 242 .. J period Vukman \noindentw-T h-h sub242 e-e\ s-nquad c-uJ e s-r . e-tVukman i-l sub a-t t-u sub i-t i-o o-n open parenthesis to the power of y-one nine-x closing parenthesis o-f sub o-r ll o-y w-i s-n sub period N-one sub e-five parenright-x g-t w-v e-e p-b r o-e c-v e-a t-s e-h sub e r-f e l-parenleft sub one-a t-nine\ centerline sub i-parenright{ $ w−T on h−h { e−e s−n c−u e } s−r e−t i−l { a−t t−u { i−t i−o o−n }} ( ˆopen{ y− parenthesisone nine 25− closingx } parenthesis) o−f open{ o− parenthesisr }$ l l x closing $ o−y parenthesis w−i comma s−n { x brackright. } N−one y x 2{ =e 0− xfive comma in parenright −x g−tPut} ngw− ..v y T xe− closinge p parenthesis−b $ r fo $ r o y− inte h cea−v b o ve e− reatia onwe t−s .. ob e− ah .. n{ e } r−f $ e $ l−p a r e n l e f t { one−a t−nineopen parenthesis{ i−parenright 26 closing parenthesis on }}$ T} open parenthesis x closing parenthesis comma x brackright yT open parenthesis x closing parenthesis x to the power of 2 = x y in \ [(25)(x),x]yx2=0x,Rg .. htm ui pli ati n of open parenthesis 25 closing parenthesis by T x closing parenthesis\ in g ivs\ ] 242 J . Vukman open parenthesis 27 closing parenthesis x brackleft T open parenthesis x closing parenthesis comma x brackright yx 2 T open w − T h − h s − re − ti − l (y−onenine−x)o − f ll o − yw − is − n N − one w − ve − ep − b r o − ec − ve − at − se − h r − f e l − parenleft e−es−nc−ue a−tt−uiparenthesis−ti−oo−n x closing parenthesiso−r = 0 comma x comma. y ine−fiveparenright R period −xg−t e one−at−ninei−parenrighton \ centerlineSubtr tn .. g{Put open ng parenthesis\quad 27$y closing T parenthesis x )$ fro m for .. 2 .. 6 $y$ closing parenthesis intheabovereationwe we o bta n x brackleft T \quad ob a \quad n } (25) (x), x]yx2 = 0 x, ∈ c to the power of a .. n .. m \ [(x T x 26)T(x) to the power of brackrightPut ng y openyT ,xx) parenthesis fo r y int ]yT(x)xˆ h x ea sub b o brackright ve reati onwe brackleft ob to a{ the n2 power} = of T x open y parenthesis\ in \ ] x x = 0 comma x sub comma in A c-c r-o sub d ng .. to open parenthesis 9(26) closingT ( parenthesisx), x]yT (x)x one2 = c anxy ∈ \ centerline2 4 closing parenthesis{Rg \quad ithtmui x sub f ll opli w T atinof stha t .. = (25)by 0 x to the power $T of comma x y )$ in R .. givs w .. t } Rg htm ui pli ati n of ( 25 ) by T x) g ivs brackleft T open parenthesis x closing parenthesis xx = 0 comma x in R period \ [(27)x[T(x),x]yx2T(x)=0,x,From the abo e re at on .. one obta (27) x[T (x), x]yx2T (x) = 0, x, y ∈ R. y T\ openin parenthesisR. \ x] closing parenthesis comma y brackright x plus brackleft T open parenthesis x x = Rig h-t ..Subtr mu titn l-p g ic ( tion 27 ) .. fro of m hea 2b .. 6 overe ) we lato o bta nby n x brackleft[T ca n T open m parenthesis x parenright-comma x brackright giv sbe cause .. of open parenthesis 24 closing parenthesis \noindent Subtr tn \quad g ( 27 )]y f r o m T\quad 2 \quad 6)weobtan $x [ T$ T open parenthesis x brackleft-y-TxT parenleft-plus x ( x] [ y-parenright( x x = to the 0 power, x, of∈ x-comma x-parenright brackright equal-y sub brackleft$ c ˆ T-T{ a open}$ parenthesis\quad n \ x-parenleftquad m parenright-parenright-comma x to the power of plus-brackright = x T-zero sub comma A c − cr − o ng to ( 9 ) one c an open parenthesis y comma-x-parenright comma y x-elementd sub comma comma-element R 2 4 ) it x ll o w T stha t = 0 x,y ∈ R w t \ [I-w-n x Ti-o-h to x the ˆ{ power] y of} c-r sub( d-h x f e-i-m-r{ ] } tp l-o[ ˆ i{ e-pT to} the( power x of r-s x o-parenleft = 0 eight-v , x e-parenright{ , }\ subin period\ ] t O-h u a-r n-b o-e sub x-v r t-e sub l-a s-a-k i-t o-s nt o-w ep n-r-o e-e-v sub d t-h to the power of e-h e r-r e-e sub l-l t-a-a t o-o-n ns b [T (x) xx = 0, x ∈ R. l-e sub o w \ centerlineLine 1 x A open{A parenthesis$ c−c r x− commao { d yFrom} closing$ ngthe parenthesis abo\quad e reto at x on = ( 0 9 comma one ) one obta x c in an R comma} Line 2 open parenthesis 3 sub closing parenthesis to the power of 0 brackleft A sub open parenthesis x comma closing parenthesis comma x sub brackright = 0 comma x sub\ centerline y in R to the{2 power 4 ) of i t comma $ x { f }$ llowT (x), y]x $T$+ [T ( sthatxx = \quad $ = 0 x ˆ{ , } y \ in R $ \quad w \quad t } Rig h − t mu ti l − p ic tion of hea b overe lato nby [T (xparenright − commax] giv sbe \ [[T(x)xx=0cause of ( 24 ) ,x \ in R. \ ]

x−comma plus−brackright \ centerlineT (xbrackleft{From the−y− aboT parenleft e re− atplusy on −\parenrightquad one obta x}−parenright]equal−y[T −T (x−parenleftparenright−parenright−commax = xT −zero,(ycomma−x−parenright, yx−element, comma−elementR c−rd−h r−s e−h I − w − ni − o − h e − i − m − r tp l − o i e − p \o[T(x),y]x+[T(xx=− parenlefteight − ve − parenright.tO−h u a − rn − bo − ex−v r t − el−as−a−ki−t\o]− s nt o − w ep n − r − oe − e − vdt − h e r − re − el−lt − a − a t o − o − n ns b l − eo w xA(x, y)x = 0, x ∈ R, 0 , \ hspace ∗{\ f i l l }Rig $ h−t $ \quad(3)mu[A t(x i, ),x $] = l− 0p, $ xy i c∈ t iR o n \quad o f hea b \quad overe lato nby $[ T ( x parenright −comma x ] $ giv sbe cause \quad o f ( 24 )

\ [T ( x brackleft −y−T p a r e n l e f t −plus y−parenright ˆ{ x−comma } x−parenright ] equal−y { [ } T−T ( x−parenleft parenright −parenright −comma x ˆ{ plus−b r a c k r i g h t } = x T−zero { , } ( y comma−x−parenright , y x−element { , } comma−element R \ ]

\ centerline { $ I−w−n i−o−h ˆ{ c−r { d−h }} e−i−m−r $ tp $ l−o $ i $ e−p ˆ{ r−s } o−p a r e n l e f t eight −v e−p a r e n r i g h t { . t O−h }$ u $ a−r n−b o−e { x−v }$ r $ t−e { l−a s−a−k i−t } o−s $ nt $ o−w $ ep $ n−r−o e−e−v { d } t−h ˆ{ e−h }$ e $ r−r e−e { l−l } t−a−a $ t $ o−o−n $ ns b $ l−e { o }$ w }

\ [ \ begin { a l i g n e d } xA(x,y)x=0,x \ in R, \\ ( 3 ˆ{ 0 } { ) } [A { ( x } ,){ , } x { ] } = 0 , x { y }\ in R ˆ{ , }\end{ a l i g n e d }\ ] Centralizers on semiprime rings .. 243 Centralizerswhere A open onparenthesis semiprime x comma rings y closing\quad parenthesis243 stands for T open parenthesis xy plus yx closing parenthesis minus T openwhere parenthesis $A y closing( x parenthesis , y x minus )$ xT standsfor open parenthesis $T y closing ( parenthesis xy + period yx .. ) Let− us firstT prove ( rela y hyphen ) x − tionxT open ( parenthesis y ) 29 closing . $ \ parenthesisquad Let period us first The substitution prove rela xy plus− yx for y in open parenthesis 1 closing parenthesis gives \noindentEquation:tion open parenthesis ( 29 ) . 3 The 1 closing substitution parenthesis .. $xy T open parenthesis+ yx$ x tofor the $y$ power of in2 yx ( plus 1 ) xyx gives to the power of 2 closing parenthesis = xT open parenthesis xy plus yx closing parenthesis x comma x comma y in R period \ beginOn the{ a other l i g n ∗} hand we obtain by putting z = x to the power of 2 in open parenthesis 1 0 closing parenthesis \ tagEquation:∗{$ ( open 3 parenthesis 1 ) $ 32} T closing ( parenthesis x ˆ{ 2 ..} T openyx parenthesis + xyx x ˆto{ the2 power} ) of 2 = yx plus xT xyx ( to the xy power + of 2 yxclosing ) Centralizers on semiprime rings 243 where A(x, y) stands for T (xy + yx) − T (y)x − xT (y). Let parenthesisx , x = xT , open y parenthesis\ in R. y closing parenthesis x to the power of 2 plus x to the power of 2 T open parenthesis y closing us first prove rela - parenthesis\end{ a l i g xn ∗} comma x comma y in R period tion ( 29 ) . The substitution xy + yx for y in ( 1 ) gives By comparing open parenthesis 3 1 closing parenthesis and open parenthesis 32 closing parenthesis we arrive at open parenthesis 29\noindent closing parenthesisOn the period other .. handFrom open we obtain parenthesis by 29 putting closing parenthesis $ z = one obtains x ˆ{ 2 open}$ parenthesis in ( 1 0 see ) how open parenthesis 20 closing parenthesisT (x2 wasyx + obtainedxyx2) = fromxT (xy open+ yx parenthesis)x, x, y ∈ 9R. closing parenthesis closing(31) parenthesis \ beginxA open{ a l i parenthesis g n ∗} x comma y closing parenthesis z plus xA open parenthesis z comma y closing parenthesis x plus zA open \ tag ∗{$ ( 32 ) $} T ( x ˆ{ 2 } yx + xyx ˆ{ 2 } ) = xT ( y ) x ˆ{ 2 } parenthesisOn x comma the other y closing hand weparenthesis obtain by x putting = 0 commaz = x x2 commain ( 1 0 y ) comma z in R period +Right x ˆ{ multiplication2 } T (of the y above ) relation x , by A x open , parenthesis y \ in x commaR. y closing parenthesis x gives because of open parenthesis\end{ a l i g 29n ∗} closing parenthesis Equation: parenleft-parenleft to theT (x power2yx + ofxyx three-three2) = xT (y) three-fourx2 + x2T ( parenright-parenrighty)x, x, y ∈ R. .. x-brackleft A-T(32) parenleft-parenleft x-x\noindent sub parenright-commaBy comparing y-y closing ( 3 1 parenthesis ) and ( z 32 plus-A ) we brackleft-parenleft arrive at ( 29 T-x ) sub . comma\quad y-yFrom parenright-parenright ( 29 ) one obtains comma-x ( see how =( sub 20 x ) equal-zero wasBy comparing obtained sub comma ( 3 from 1 x ) andcomma ( ( 9 32 )y-x ) )we sub arrive comma at ( y-z 29 element-element ) . From ( 29 )R-R one sub obtains period ( see how ( 20 ) was P-L subobtained u-e t-t i-u-t from n-s ( 9p-g ) ) to the power of x-r y-o v plus-e r-y x-e l f-a o-t r-i sub o y-n n-parenleft 3 t-zero parenright-h sub period-e\ [xA(x a-T b-h sub e-o ,y)z+xA(z l-e i r-n e-e l-a a-r i-t z-i a-o n-t i o-a n-n ,y)x+zA(x o f-u s-parenleft to the power of n-eight closing ,y)x parenthesis =g-parenleft 0 ,i v-eight x parenright-e , y ,w-sxA z(x, y)z\ in+ xA(z,R. y)x + zA\ ](x, y)x = 0, x, y, z ∈ R. Equation:Right y to multiplication the power of x of brackright the above plus relation .. = to by theA(x, power y)x gives of x brackleft because ofopen ( 29 parenthesis ) closing parenthesis plus brackleft open parenthesis closing parenthesis x plus x y comma plus x y T = x R \noindentAcc o di ngtRight open multiplicationparenthesis 34 closing of parenthesis the above o ne relation ac n r-e l-p by a c e$A .. T open ( parenthesis x , x y comma ) y x brackright $ gives by ..because of ( 29 ) minus-brackleftx − brackleftA T open parenthesis− T parenleft y comma-parenright− parenleftx − xx brackright-i nth ey a− by) v-ozplus ere− aAbrackleft t-i n-o sub period− parenleftT W − x y − yparenright − parenrightcomma − x = equal − zero x, y − x y − zelement − elementR − R \ begin { a l i g n ∗} parenright−comma , x , , . brackleft T open parenthesis xy plus yx closingparenleft parenthesis− parenleft minusthree comma−threethree T y−four closingparenright parenthesis− parenright x T minus open parenthesis sub\ tag x∗{ T$ open p a parenthesis r e n l e f t −p y a commar e n l e f tx brackright ˆ{ three =−thre open e parenthesis three − 0fo period ur } .. Thparenright e .. o o f f− r-eparenright sub l $} x−b r a c k l e f t A−T p a r e n l e f t −p a r e n l e f t x−x { parenright −comma } y−y ) z plus−A b r a c k l e f t −p a r e n l e f t A open parenthesisP − L xi comma− u − tn y− closingsp − g parenthesisx−ry−o v plus r x− perioder − yx .. Th− e uswl f − t ehaao − vetr ..− openi parenthesisn − parenleft .. m3t ..− a .. e .. t p .. i to T−x { , } y−u−yet−t parenright −parenright comma−x = { x } equaloy−−nzero { , } x , y−x { , } the power ofzeroparenright l e .. t .. r p ..− eh .. n .. Aa x− ..T y b − h l − e i r − ne − el − aa − ri − tz − ia − on − t i o − an − n o y−z element−elementperiod R−R−e { . } e−o Equation:f − parenleft-parenleft-parenleftus − parenleftn−eight)g − parenleft three-three-threei v − eightparenright parenright-parenright-parenright− ew − s .. A x y-x-A A-parenright x-parenleft x-parenleft\end{ a l i g comma-x-z n ∗} closing parenthesis comma equal-equal-y zero-zero x-comma-comma = x-x 0 y sub y element-element-x sub comma period-period in R comma 6 open parenthesis A R w-O f-h e-c sub n u-c e r-s sub t f-w o-e-l sub o-l h v-s t-e h-a sub t-s-a o \noindent $ P−L { u−e t−t } i−=ux−[t ( n )−s + p [−g ( ˆ ){ xx−r+ y−o }$ v $ plus−eyx]+ r−y x−e $ l $ f−a o−t r−i { o y−n } n−parenleft 3 t−zero parenright −h { period−e } a−T b−h { e−o } x y , + x y T = x R l−e $ i $ r−n e−e l−a a−r i−t z−i a−o n−t $ i $ o−a n−n $ o $ f−u s−p a r e n l e f t ˆ{ n−e i g h t } ) g−parenleftAcc o di ngt $ ( i 34 $ ) v o− neeight ac n r − parenrightel − p a c− eeT ( wx,−s y] $ by minus − brackleftT (ycomma − parenrightxbrackright − i nth e a b v − o ere a t − in − o. W \ begin { a l i g n ∗} [T (xy + yx)−, T y)xT − (xT (y, x] = (0. Th e o o f f r − el A(x, y)rx. Th usw t eha ve ( m \ tag ∗{$ ya ˆ{ ex t p ] i}l e+ t $} r= p ˆ{ ex n} A[ x (y ) + [ ( ) x + \\ x y , + x y T = x R \end{ a l i g n ∗}

Axy − x − AA − parenrightx − parenleftx − parenleftcomma − x − z), equal − equal − yzero − zerox − comma − comma = x − x0yyelement − element − x, period − period ∈ R, \noindent Accoparenleft di ngt− parenleft ( 34 )− oparenleftthree ne acn $− rthree−e− lthreeparenright−p $ a c e −\quadparenright$ T− parenright ( x , y ] $ by \quad $ minus−brackleft T ( y comma−parenright x brackright −i $ nth e a b $ v−o $ 6 ( AR e r e a $ t−i n−o { . }$ W w − Of − he − cnu − cer − stf − wo − e − lo−lhv − st − eh − at−s−ao \noindent $ [ T ( xy + yx ) − , T y ) x T − ( { x } T ( y , x ] = ( 0 . $ \quad Th e \quad o o f f $ r−e { l }$ $ A ( x , y ) r { x } . $ \quad Th usw t eha ve \quad ( \quad m \quad a \quad e \quad t p \quad $ i ˆ{ l }$ e \quad t \quad r p \quad e \quad n \quad $ A x $ \quad y

\ begin { a l i g n ∗} \ tag ∗{$ p a r e n l e f t −p a r e n l e f t −parenleft three −three −three parenright −parenright −parenright $} A x y−x−AA−parenright x−p a r e n l e f t x−parenleft comma−x−z ) , equal−equal−y zero−zero x−comma−comma = x−x 0 y { y } element−element−x { , } period−period \ in R, \\ 6 (AR \\ w−O f−h e−c { n } u−c e r−s { t } f−w o−e−l { o−l h } v−s t−e h−a { t−s−a o } \end{ a l i g n ∗}