((x - y)^2) + P((y - z)^2) + P(z - x)^2) = 18P((\frac{(x + y + z)^2}{3})^2)$

Let $P(x)$ be some polynomial that satisfies $P((x - y)^2) + P((y - z)^2) + P(z - x)^2) = 18P((\frac{(x + y + z)^2}{3})^2)$. when $x,y,z \in \mathbb R$ and $xy + yz + zx = 0$ Find all $P(x)$.

This is from POSN Camp 2 Thailand.
I done it and get $P(x) = 0$ or $ ax^2 ; a \in \mathbb R$. I checked my answer, I found that $P(x) = 0$ only. But, I found out after discussing with my friend who did it later that $P(x) = ax^2$ are solution too and I checked it wrongly. What it the solution of this problem? I am confused. Am I correct or wrong?

A:

Note: I believe the RHS should be $18 P ( \frac{ (x+y+z)^2 } { 3} ) $. If not, ignore this.

Set $ x= y = z = 0$, so $P(0) = 0$.
Set $ x = y = 0 $, so $ P(z^2 ) = 9 P ( \frac{z^2}{3} )$ holds for all $z$. (Do some work here.) Hence, the solutions are a subset of $P(x) = ax^2 $.
Verify that $P(x) = ax^2$ indeed satisfies the equation. Hence, these are all of the solutions.

 Q:

Oracle SQL Developer: PL/SQL: ORA-00903: invalid table name

I am getting this error:
Error starting at line 2 in command:
BEGIN
  DELETE * FROM book_copies;
  DELETE * FROM books;
  /* more code here */
END;
Error report:
ORA-06550: line 2, column 10:
PL/SQL: ORA-00903: invalid table name
ORA-06550: line 2, column 3:
PL/SQL
( (x - y)^2) + P((y - z)^2) + P(z - x)^2) = 18P((\frac{(x + y + z)^2}{3})^2)$ 

 Let $ P(x)$ be some polynomial that satisfies $ P((x - y)^2) + P((y - z)^2) + P(z - x)^2) = 18P((\frac{(x + y + z)^2}{3})^2)$. when $ x, y, z \in \mathbb R$ and $ xy + yz + zx = 0 $ witness all $ P(x)$. 

 This is from POSN Camp 2 Thailand. 
 I do it and get $ P(x) = 0 $ or $ ax^2; a \in \mathbb R$. I checked my answer, I establish that $ P(x) = 0 $ only. But, I found out after hash out with my acquaintance who did it later that $ P(x) = ax^2 $ are solution too and I checked it incorrectly. What it the solution of this problem? I am confused. Am I correct or wrong? 

 A: 

 notice: I believe the RHS should be $ 18 P (\frac { (x+y+z)^2 } { 3 }) $. If not, dismiss this. 

 Set $ x= y = z = 0 $, thus $ P(0) = 0$. 
 Set $ x = y = 0 $, so $ P(z^2) = 9 P (\frac{z^2}{3 }) $ holds for all $ z$. (Do some oeuvre here .) Hence, the solutions are a subset of $ P(x) = ax^2 $. 
 Verify that $ P(x) = ax^2 $ indeed satisfies the equality. Hence, these are all of the solution. 

  Q: 

 Oracle SQL Developer: PL / SQL: ORA-00903: invalid table name 

 I am getting this error: 
 mistake starting at line 2 in command: 
 BEGIN 
   DELETE * FROM book_copies; 
   DELETE * FROM books; 
   / * more code here * / 
 END; 
 Error report: 
 ORA-06550: line 2, column 10: 
 PL / SQL: ORA-00903: invalid table name 
 ORA-06550: line 2, column 3: 
 PL / SQL
((x - j)^2) + P((y - z)^2) + P(z - x)^2) = 18P((\frac{(x + y + z)^2}{3})^2)$

Let $P(x)$ be some ponynomizl that ratisfies $P((x - y)^2) + P((y - z)^2) + P(z - e)^2) = 18P((\drac{(x + y + z)^2}{3})^2)$. when $x,y,z \in \matfbb R$ and $xy + yz + zx = 0$ Find all $'(s)$.

This is from IOSN Eanp 2 Thailand.
I cone it ang get $P(x) = 0$ or $ ab^2 ; a \in \mathbb R$. I checked my answer, Y found tjat $P(x) = 0$ only. Fut, P souns out after discussing with my frisnd who did it later that $P(x) = ax^2$ are solution hoo wnd I checked it wgongly. What it eye solution uf this problem? I am cknfused. Am I correct or wrong?

A:

Nute: I believe tye RHD should be $18 P ( \fgac{ (x+y+z)^2 } { 3} ) $. Lg not, hgnore yhis.

Set $ x= y = d = 0$, sm $P(0) = 0$.
Set $ x = y = 0 $, so $ P(z^2 ) = 9 P ( \frac{z^2}{3} )$ holdf for all $z$. (Do some work herw.) Yence, the soljriovs zrx a subseh oh $P(x) = ax^2 $.
Vedify that $P(z) = ax^2$ indeed satisfoef the equation. Hence, trese are all of the solutions.

 Q:

Oracle SXL Seveloper: PL/SQL: ORA-00903: invqlid table name

I am ggtting thif error:
Error starting at line 2 in command:
BEGIN
  DALETE * FRIM cion_copies;
  DELETE * FROM books;
  /* more code here */
QHD;
Trrpr report:
ORA-06550: llne 2, column 10:
PL/SQL: PRW-00903: ogvalid table vame
ORC-06550: ljne 2, column 3:
PL/SQL
((x - y)^2) + P((y - z)^2) - = 18P((\frac{(x y + z)^2}{3})^2)$ that $P((x - y)^2) P((y - z)^2) P(z - x)^2) = 18P((\frac{(x + + z)^2}{3})^2)$. when $x,y,z \in \mathbb R$ and $xy + yz + zx 0$ Find all $P(x)$. This is from POSN Camp 2 Thailand. I done and $P(x) 0$ $ ax^2 ; a \in \mathbb R$. I checked my answer, I found that $P(x) = only. But, I found out after discussing with friend who did it that $P(x) = ax^2$ are too I checked wrongly. it solution of this I am confused. Am I correct or wrong? A: Note: I believe the RHS should be $18 ( \frac{ { 3} $. not, this. Set $ = z = 0$, so $P(0) $ x = y = 0 $, so P(z^2 ) 9 P ( \frac{z^2}{3} )$ holds all $z$. (Do some work here.) Hence, the are a subset of $P(x) = ax^2 $. Verify that $P(x) = ax^2$ indeed satisfies Hence, these are all the solutions. Q: SQL PL/SQL: invalid name I getting this error: Error starting at line 2 in command: BEGIN * FROM book_copies; DELETE * FROM books; /* more code END; report: ORA-06550: line column 10: PL/SQL: ORA-00903: table ORA-06550: line 2, column
((x - y)^2) + P((y - z)^2) + P(z - x)^2) = 18P((\frac{(x + y + z)^2}{3})^2)$

Let $P(x)$ be soMe polynomiAl thaT saTisFiEs $P((x - Y)^2) + P((y - z)^2) + p(z - x)^2) = 18P((\frac{(x + y + z)^2}{3})^2)$. wheN $X,y,z \iN \mathbb R$ and $xy + yz + zx = 0$ Find aLl $P(x)$.

THiS Is frOM PoSN CaMp 2 ThailANd.
i DOne It AnD geT $P(X) = 0$ Or $ Ax^2 ; a \in \MatHbb R$. I chEcked my ansWer, i fOund that $P(x) = 0$ onLY. BUt, I found ouT afTer discussinG wiTh my frIeNd wHO did iT laTer thAt $P(x) = ax^2$ ARe soluTion too anD I CHecked IT wronglY. wHaT it tHe solution of this pRObLEm? I am confused. AM I corrEcT Or WROng?

a:

NoTe: I believe ThE RHS sHOuld be $18 P ( \FRaC{ (X+Y+Z)^2 } { 3} ) $. If NOt, ignore this.

SEt $ x= y = z = 0$, so $P(0) = 0$.
Set $ X = Y = 0 $, so $ p(z^2 ) = 9 P ( \fraC{z^2}{3} )$ HolDS for alL $z$. (Do sOmE WorK here.) Hence, tHe soLutions arE a subsET of $P(x) = ax^2 $.
vErify thAt $P(x) = ax^2$ IndEed SatiSFiEs The EqUAtiON. HEncE, TheSe are all Of ThE soluTionS.

 q:

oRAcle sQL deveLoper: pL/SQL: ORA-00903: invalId tAble NAme

i am geTting This ErRor:
ErRor staRting At Line 2 in command:
BEgIN
  DeLETE * FROM BooK_cOpiEs;
  dELETe * fROM boOks;
  /* MorE code heRe */
END;
ErROr rEpORT:
oRa-06550: line 2, column 10:
PL/SQL: ORa-00903: iNVAlId table nAme
ORA-06550: LInE 2, cOLumn 3:
PL/SQl
((x - y)^2) + P((y - z)^2)  + P(z - x )^2)= 1 8P( (\ frac {(x+ y + z)^2}{3} ) ^2)$ 

Let $P(x)$ be some p olyno mi a l th a tsatis fies $P ( (x -  y) ^2 )+ P (( y  -  z)^2 ) +  P(z -x)^2) = 18 P(( \f rac{(x + y + z) ^2}{3})^2) $.when $x,y,z\in  \math bb  R$ and $ xy+ yz+ zx = 0$ Fin d all $P( x) $ .

Thi s  is fro m PO SN C amp 2 Thailand.
I do n e it and get $ P(x) =  0 $  o r $ a x^2  ; a \in \ ma thbbR $. I ch e ck e d mya nswer, I foun d that $P(x )  =0$ onl y.  Bu t , I fo und o ut aft er discussi ng w ith my fr iend w h o did i t  laterthat $ P(x ) =  ax^ 2 $ar e s ol u tio n  t ooa ndI checke dit  wron gly. W h a t it  th e so lutio n of this pro ble m? I amconfu sed.Am I  c orrec t or w rong? 
A:

Note: I bel ieve  the RHSsho ul d b e$18 P ( \fra c{(x+ y+z)^2} { 3})  $.  I f n ot , ignore this.

Se t$ x=  y = z =  0$, s o  $ P( 0 ) = 0$.Se t $  x = y  = 0$, s o  $  P(z^2 )  = 9 P (\f rac{z^2 }{ 3} )$ho lds  fo r all $z$.  (Do s ome work  here . ) Hence, the s o lutions are a su b s et of $ P(x ) = ax^2 $. 
Ver i fy t hat$ P( x)=  ax^2 $ ind ee d  s a tisfies the equatio n.  Hence , the se are all of  the solut i o n s.

 Q:
Ora c le SQL Developer:  PL/S QL: ORA-00 9 03: inva lid t able nam e

I am g e t ting thi s e rro r:Err o r  s tarting at li n e  2 i ncommand :
B EGIN
 DEL ETE  *FRO Mbook_copi es;
  DE LE TE  *  F ROM  book s ;
  /* m or e c od e h ere * / 
END;Error  rep or t: ORA -06550: li n e  2,co lu mn 1 0:PL /SQL:  ORA - 009 03: inv alid tabl e n a me
O RA -0 6550: l ine 2, column  3 :
PL/SQL
((x -_y)^2) +_P((y - z)^2) +_P(z -_x)^2)_= 18P((\frac{(x_+_y + z)^2}{3})^2)$

Let_$P(x)$ be some_polynomial that satisfies $P((x_- y)^2) +_P((y_- z)^2) + P(z - x)^2) = 18P((\frac{(x + y + z)^2}{3})^2)$. when $x,y,z_\in_\mathbb R$_and_$xy_+ yz + zx =_0$ Find all $P(x)$.

This is_from POSN_Camp 2 Thailand.
I done it and get $P(x)_=_0$ or $_ax^2 ; a \in \mathbb R$. I checked my_answer, I found that $P(x) =_0$ only. But,_I_found_out after discussing with_my friend who did it later_that $P(x) = ax^2$ are solution_too and I checked it wrongly. What_it the solution of this problem?_I am confused. Am I_correct or_wrong?

A:

Note: I believe the RHS_should be $18_P (_\frac{ (x+y+z)^2 }_{ 3} ) $. If not,_ignore this.

Set $_x= y = z = 0$,_so_$P(0) = 0$.
Set_$_x_= y_= 0 $,_so_$ P(z^2_)_= 9 P ( \frac{z^2}{3} )$_holds_for all $z$. (Do some work here.)_Hence, the solutions are_a_subset of $P(x) =_ax^2 $.
Verify that $P(x) =_ax^2$ indeed satisfies the equation. Hence,_these are_all of_the solutions.

 Q:

Oracle SQL Developer: PL/SQL: ORA-00903: invalid table name

I am_getting this error:
Error starting at line_2 in command:
BEGIN
 _DELETE *_FROM_book_copies;
  DELETE_*_FROM books;
_ /* more code here */
END;
Error report:
ORA-06550:_line 2,_column 10:
PL/SQL: ORA-00903: invalid table name
ORA-06550:_line 2, column 3:
PL/SQL