Question

Factorize
$\begin{array}{cc}& x^{12}-y^{12}\\ & x^9-y^9\end{array}$

Answer 1

$x^{12}-y^{12}$

$={\left(x^6\right)}^2-{\left(y^6\right)}^2$

$=(x^6+y^6)(x^6-y^6)$

$x^6-y^6$

$={\left(x^3\right)}^2-{\left(y^3\right)}^2$

$=(x^3+y^3)(x^3-y^3)$

$=(x+y)(x^3-y^3)(x^2-xy+y^2)$

$=(x+y)(x-y)(x^2-xy+y^2)(x^2+xy+y^2)$

$=(x+y)(x-y)(x^6+y^6)(x^2-xy+y^2)(x^2+xy+y^2)$

$\therefore x^{12}-y^{12}=(x+y)(x-y)(x^6+y^6)(x^2-xy+y^2)(x^2+xy+y^2)$

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$x^9-y^9$

$Applyfactoringrule:x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\cdots +x{y}^{n-2}{y}^{n-1})$

$x^9-y^9=(x-y)(x^8+x^{7}y+x^6{y}^2+x^5{y}^3+x^4{y}^4+x^3{y}^5+x^2{y}^6+x{y}^7+y^8)$

$\therefore x^9-y^9:(x-y)(x^8+x^{7}y+x^6{y}^2+x^5{y}^3+x^4{y}^4+x^3{y}^5+x^2{y}^6+x{y}^7+y^8)$