Question

Factorise the following: $a^{12}{x}^4-a^4{x}^{12}$

Answer 1

Using $a^2-b^2=(a+b)(a-b)$ at each step

Given:$a^{12}{x}^4-a^4{x}^{12}$

$=a^4{x}^4(a^8-x^8)$, factor out common factor $a^4{x}^4$

$=a^4{x}^4\left[\right(a^4{)}^2-\left(x^4{)}^2\right]$

$=a^4{x}^4(a^4+x^4)(a^4-x^4)$

$=a^4{x}^4(a^4+x^4)\left[\right(a^2{)}^2-\left(x^2{)}^2\right]$

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

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