Question

Factorise $x^6-y^6$ Solve This.

Answer 1

Step 1: Factorize the given polynomial

$x^6-y^6={\left(x^2\right)}^3-{\left(y^2\right)}^3$

Using the identity $a^3-b^3=(a-b)(a^2+b^2+ab)$

$x^6-y^6=(x^2-y^2)(x^4+y^4+x^2\times y^2)$

Using the identity $a^2-b^2=(a-b)(a+b)$

$x^6-y^6=(x-y)(x+y)(x^4+y^4+x^2{y}^2)$

Step 2: Further simplify the factor

adding and subtracting $x^2{y}^2$

$$\begin{gathered} x^6-y^6=(x-y)(x+y)(x^4+y^4+x^2{y}^2+x^2{y}^2-x^2{y}^2) \\ {x}^6-y^6=(x-y)(x+y)(x^4+y^4+2x^2{y}^2-x^2{y}^2) \end{gathered}$$

using identity ${(a+b)}^2=a^2+b^2+2ab$

$$\begin{gathered} x^6-y^6=(x-y)(x+y)(x^4+y^4+2x^2{y}^2-x^2{y}^2) \\ {x}^6-y^6=(x-y)(x+y)\left[{(x^2+y^2)}^2-x^2{y}^2\right] \end{gathered}$$

again using identity $a^2-b^2=(a-b)(a+b)$

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

Hence the factorized form of the polynomial $x^6-y^6$ is $(x-y)(x+y)(x^2+y^2-xy)$$(x^2+y^2+xy)$.