The problem is to verify the identity: $sin^6x + cos^6x = 1 - 3sin^2xcos^2x$

1. Problem Description

The problem is to verify the identity:

$sin^6x + cos^6x = 1 - 3sin^2xcos^2x$

2. Solution Steps

We start from the left-hand side (LHS) and try to derive the right-hand side (RHS).

Recall the identity:

$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$

We can rewrite the LHS as:

$sin^6x + cos^6x = (sin^2x)^3 + (cos^2x)^3$

Let $a = sin^2x$ and $b = cos^2x$. Then, $a+b = sin^2x + cos^2x = 1$.

Using the identity, we have:

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

Since $sin^2x + cos^2x = 1$, we have:

$(sin^2x)^3 + (cos^2x)^3 = (sin^4x - sin^2xcos^2x + cos^4x)$

Now, we can rewrite $sin^4x + cos^4x$ as follows:

$sin^4x + cos^4x = (sin^2x + cos^2x)^2 - 2sin^2xcos^2x$

Since $sin^2x + cos^2x = 1$,

$sin^4x + cos^4x = 1 - 2sin^2xcos^2x$

Therefore,

$sin^4x - sin^2xcos^2x + cos^4x = (1 - 2sin^2xcos^2x) - sin^2xcos^2x = 1 - 3sin^2xcos^2x$

Hence, $sin^6x + cos^6x = 1 - 3sin^2xcos^2x$

3. Final Answer

$sin^6x + cos^6x = 1 - 3sin^2xcos^2x$