Question

Factorise x^4-(x-z)^4​

Answer 1

$\textbf{Given:}$

$x^4-(x-z)^4$

$=(x^2)^2-((x-z)^2)^2$

$\text{using}$

$\boxed{\bf\;a^2-b^2=(a-b)(a+b)}$

$=(x^2-(x-z)^2)(x^2+(x-z)^2)$

$\text{using}$

$\boxed{\bf\;(a-b)^2=a^2+b^2-2ab}$

$=(x^2-(x^2+z^2-2xz))(x^2+(x^2+z^2-2xz))$

$=(x^2-x^2-z^2+2xz)(2x^2+z^2-2xz)$

$=(-z^2+2xz)(2x^2+z^2-2xz)$

$=(2xz-z^2)(2x^2+z^2-2xz)$

$=z(2x-z)(2x^2+z^2-2xz)$

$\implies\bf\;x^4-(x-z)^4=z(2x-z)(2x^2+z^2-2xz)$

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