# Ploynomial...

• January 25th 2011, 05:43 AM
breitling
Ploynomial...
Not sure if im going mad but struggling with this (Thinking)
(a) Write down the polynomial P in x, y such that:
x^6 y^6 = (x^2 y^2)*P.

• January 25th 2011, 05:59 AM
FernandoRevilla
Hint :

Denote $t=x^2$ and $a=y^2$ then,

$x^6-y^6=t^3-a^3$

so,

$a$ is a root of $q(t)=t^3-a^3$

and you can decompose it using Ruffini's rule.

Fernando Revilla
• January 25th 2011, 06:10 AM
Soroban
Hello, breitling!

Quote:

$\text{Write down the polynomial }P\text{ in }x, y\text{ such that:}$

. . $x^6 - y^6 \:=\: (x^2 - y^2)\cdot P$

Don't you recognize that difference of cubes?

. . . . $a^3 - b^3 \:=\:(a-b)(a^2 + ab + b^2)$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The very least you can do is some Long Division . . .

. . $\begin{array}{cccccccccc} &&&& x^4 & + & x^2y^2 &+& y^4 \\ && -- & -- & -- & -- & -- & -- & -- \\ x^2-y^2 & | & x^6 &&&&& - & y^6 \\ && x^6 &-& x^4y^2 \\ && -- & -- & -- \\ &&&& x^4y^2 \\ &&&& x^4y^2 &-& x^2y^4 \\ &&&& -- & -- & -- \\ &&&&&& x^2y^4 &-& y^6 \\ &&&&&& x^2y^4 &-& y^6 \\ &&&&&& -- & -- & -- \end{array}$

• January 25th 2011, 07:32 AM
Prove It
A better solution:

$\displaystyle x^6 - y^6 = (x^2)^3 - (y^2)^3$

$\displaystyle = (x^2 - y^2)[(x^2)^2 + x^2y^2 + (y^2)^2]$ by the Difference of Two Cubes rule...

$\displaystyle = (x^2 - y^2)(x^4 + x^2y^2 + y^4)$.

So what is $\displaystyle P$?
• January 31st 2011, 05:03 AM
breitling
The question actually asked x^6-y^6=(x^2+y^2)*p got the sign wrong.
• January 31st 2011, 05:25 AM
HallsofIvy
Quote:

Originally Posted by breitling
The question actually asked x^6-y^6=(x^2+y^2)*p got the sign wrong.

Then do exactly what people suggested you do before!