Verify the trigonometric identity: $\sin^6 x + \cos^6 x = 1 - 3\sin^2 x \cos^2 x$.

1. Problem Description

Verify the trigonometric identity: $\sin^6 x + \cos^6 x = 1 - 3\sin^2 x \cos^2 x$.

2. Solution Steps

We start with the left-hand side (LHS) of the equation and manipulate it to obtain the right-hand side (RHS).

We know that $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$. Let $a = \sin^2 x$ and $b = \cos^2 x$. Then, $a^3 + b^3 = \sin^6 x + \cos^6 x$.

So we have:

$\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)((\sin^2 x)^2 - \sin^2 x \cos^2 x + (\cos^2 x)^2)$.

We know that $\sin^2 x + \cos^2 x = 1$.

Therefore,

$\sin^6 x + \cos^6 x = 1 \cdot (\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x)$.

$\sin^6 x + \cos^6 x = \sin^4 x - \sin^2 x \cos^2 x + \cos^4 x$.

Now, we add and subtract $3\sin^2 x \cos^2 x$:

$\sin^6 x + \cos^6 x = \sin^4 x + 2\sin^2 x \cos^2 x + \cos^4 x - 3\sin^2 x \cos^2 x$.

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

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

$\sin^6 x + \cos^6 x = (1)^2 - 3\sin^2 x \cos^2 x$.

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

3. Final Answer

The identity is verified: $\sin^6 x + \cos^6 x = 1 - 3\sin^2 x \cos^2 x$.