Hi, can someone solve this for me? Thanks :)

Factorise fully:

(a^2-b^2)^2-(a-b)^4(a2−b2)2−(a−b)4 =?

How do I work through this?

The solution is (4ab)(a-b)^2(4ab)(a−b)2

1 Answer
Jim G.
Apr 28, 2018

"see explanation"see explanation

Explanation:

a^2-b^2" is a "color(blue)"difference of squares"a2−b2 is a difference of squares

•color(white)(x)a^2-b^2=(a-b)(a+b)∙xa2−b2=(a−b)(a+b)

=((a-b)(a+b))^2-(a-b)^4=((a−b)(a+b))2−(a−b)4

=(a-b)^2(a+b)^2-(a-b)^4=(a−b)2(a+b)2−(a−b)4

"take out a "color(blue)"common factor "(a-b)^2take out a common factor (a−b)2

=(a-b)^2((a+b)^2-(a-b)^2)=(a−b)2((a+b)2−(a−b)2)

=(a-b)^2(a^2+2ab+b^2-(a^2-2ab+b^2))=(a−b)2(a2+2ab+b2−(a2−2ab+b2))

=(a-b)^2(cancel(a^2)+2abcancel(+b^2)cancel(-a^2)+2abcancel(-b^2))

=4ab(a-b)^2larr"as required"

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