Question

Factorize `sin^6x-cos^6x`?

Answer 1

Please see below.

Explanation:

Using identity `a^3-b^3=(a-b)(a^2+b^2-ab)`

`sin^6x−cos^6x`

= `(sin^2x)^3-(cos^2x)^3`

= `(sin^2x-cos^2x)((sin^2x)^2+(cos^2x)^2-sin^2xcos^2x)`

= `(sin^2x-cos^2x)((sin^2x+cos^2x)^2-2sin^2xcos^2x-sin^2xcos^2x)` - as `a^2+b^2=(a+b)^2-2ab`

= `(sin^2x-cos^2x)(1-3sin^2xcos^2x)`