Question

factorise x to the power 8 minus y
to the power 8

Answer 1

Step-by-step explanation:

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Answer 2

$The factors of x^{8}- y^{8} are \left(x^{4}+y^{4}\right)\left[\left(x^{2}+y^{2}\right)(x+y)(x-y)\right].$

Given:

$x^{8}-y^{8}$

To find:

Factors of $x^{8}-y^{8}$

Solution:

We can solve the above mathematical problem with the following approach.

$\begin{array}{l}\rightarrow x^{8}-y^{8} \\ We can write this as - \\=\left(x^{4}\right)^{2}-\left(y^{4}\right)^{2} [\because a^2 - b^2 = (a+b)(a-b)] \\ =\left(x^{4}+y^{4}\right)\left(x^{4}-y^{4}\right) \\=\left(x^{4}+y^{4}\right)\left[\left(x^{2}\right)^{2}-\left(y^{2}\right)^{2}\right] [\because a^2 - b^2 = (a+b)(a-b)]\\=\left(x^{4}+y^{4}\right)\left[\left(x^{2}+y^{2}\right)\left(x^{2}-y^{2}\right)\right] \\=\left(x^{4}+y^{4}\right)\left[\left(x^{2}+y^{2}\right)(x+y)(x-y)\right]\end{array}$

Therefore, the factors of $x^{8}-y^{8} are$ $\left(x^{4}+y^{4}\right)\left[\left(x^{2}+y^{2}\right)(x+y)(x-y)$.