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

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Answer 1 · 2017-01-03T07:53:32

Please see below.

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 $a^2+b^2=(a+b)^2-2ab$

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

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