# Demonstrate equality (sinx)^6 + (cosx)^6 = 1 - 3(six*cosx)^2

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### 1 Answer

You should use the following special product, such that:

`(a + b)^3 = a^3 + b^3 + 3ab(a + b)`

Considering `a = sin^2 x` and `b = cos^2 x` yields:

`(sin^2 x + sin^2 x)^3 = (sin^2 x)^3 + (cos^2 x)^3 + 3sin^2 x*cos^2 x(sin^2 x + cos^2 x)`

You need to use Pythagorean identity, such that:

`sin^2 x + sin^2 x = 1`

Replacing `1` for `sin^2 x + sin^2 x` in the expansion above, yields:

`1^3 = sin^6 x + cos^6 x + 3sin^2 x*cos^2 x*1`

Keeping `sin^6 x + cos^6 x` to one side, yields:

`1 - 3sin^2 x*cos^2 x = sin^6 x + cos^6 x`

**The last line represents exactly the identity you need to prove, hence, testing if the given expression holds, yields thatÂ `sin^6 x + cos^6 x = 1 - 3sin^2 x*cos^2 x` represents a valid statement.**

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