Question

Cos⁶x+sin⁶x=1-3sin²xcos²x

Answer 1

It can be written as

(cos²x) ³+(sin²x)³

=(cos²x+sin²x)(cos⁴x+sin⁴x-sin²xcos²x)

As we know that

(a³+b³) =(a+b) (a²+b²-ab)

Now cos⁴x+sin⁴x can be written as

=(cos²x+sin²x) - 2sin²xcos²x

=1-2sin²xcos²x

On substituting we get

(cos²x) ³+(sin²x)³

=(1)(1-2sin²xcos²x-sin²xcos²x)

=1-3sin²xcos²x

Hope it helps you :-)

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