Question

Factorise: a^2-6a^2*b^2+b^4

Answer 1

We're asked to factorise $a^4-6a^2b^2+b^4.$

We can split $6a^2b^2$ as $2a^2b^2+4a^2b^2.$

$\longrightarrow a^4-6a^2b^2+b^4=a^4-(2a^2b^2+4a^2b^2)+b^4$

$\longrightarrow a^4-6a^2b^2+b^4=a^4-2a^2b^2-4a^2b^2+b^4$

$\longrightarrow a^4-6a^2b^2+b^4=a^4-2a^2b^2+b^4-4a^2b^2$

$\longrightarrow a^4-6a^2b^2+b^4=(a^4-2a^2b^2+b^4)-4a^2b^2\quad\quad\dots(1)$

We know that,

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

Taking $x=a^2$ and $y=b^2,$

$\longrightarrow (a^2)^2-2(a^2)(b^2)+(b^2)^2=(a^2-b^2)^2$

$\longrightarrow a^4-2a^2b^2+b^4=(a^2-b^2)^2$

Then (1) becomes,

$\longrightarrow a^4-6a^2b^2+b^4=(a^2-b^2)^2-4a^2b^2$

$\longrightarrow a^4-6a^2b^2+b^4=(a^2-b^2)^2-(2ab)^2$

Since $x^2-y^2=(x+y)(x-y),$

$\longrightarrow a^4-6a^2b^2+b^4=(a^2-b^2+2ab)(a^2-b^2-2ab)$

$\longrightarrow\underline{\underline{a^4-6a^2b^2+b^4=(a^2+2ab-b^2)(a^2-2ab-b^2)}}$

Hence Factorised!