Not sure if im going mad but struggling with this
(a) Write down the polynomial P in x, y such that:
x^6 − y^6 = (x^2 − y^2)*P.
Many thanks in advance.
Hint :
Denote $\displaystyle t=x^2$ and $\displaystyle a=y^2$ then,
$\displaystyle x^6-y^6=t^3-a^3$
so,
$\displaystyle a$ is a root of $\displaystyle q(t)=t^3-a^3$
and you can decompose it using Ruffini's rule.
Fernando Revilla
Hello, breitling!
$\displaystyle \text{Write down the polynomial }P\text{ in }x, y\text{ such that:}$
. . $\displaystyle x^6 - y^6 \:=\: (x^2 - y^2)\cdot P $
Don't you recognize that difference of cubes?
. . . . $\displaystyle a^3 - b^3 \:=\:(a-b)(a^2 + ab + b^2)$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
The very least you can do is some Long Division . . .
. . $\displaystyle \begin{array}{cccccccccc} &&&& x^4 & + & x^2y^2 &+& y^4 \\ && -- & -- & -- & -- & -- & -- & -- \\ x^2-y^2 & | & x^6 &&&&& - & y^6 \\ && x^6 &-& x^4y^2 \\ && -- & -- & -- \\ &&&& x^4y^2 \\ &&&& x^4y^2 &-& x^2y^4 \\ &&&& -- & -- & -- \\ &&&&&& x^2y^4 &-& y^6 \\ &&&&&& x^2y^4 &-& y^6 \\ &&&&&& -- & -- & -- \end{array}$
A better solution:
$\displaystyle \displaystyle x^6 - y^6 = (x^2)^3 - (y^2)^3$
$\displaystyle \displaystyle = (x^2 - y^2)[(x^2)^2 + x^2y^2 + (y^2)^2]$ by the Difference of Two Cubes rule...
$\displaystyle \displaystyle = (x^2 - y^2)(x^4 + x^2y^2 + y^4)$.
So what is $\displaystyle \displaystyle P$?