It looks like you're asking to prove a trigonometric identity using the double-angle formulas. I can help with that. From the image, I see the equation:

$\sin^3 x + \cos^3 x = (\sin x + \cos x) \left( 1 - 2 \sin^2 2x \right)$

The double-angle formulae for sine and cosine are:

$\sin 2x = 2 \sin x \cos x$ $\cos 2x = \cos^2 x - \sin^2 x$

To prove this, we need to expand both sides and match terms. Let's break it down:

Left-hand side (LHS):

$\sin^3 x + \cos^3 x$ This can be factored using the sum of cubes formula: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ Thus, we have: $\sin^3 x + \cos^3 x = (\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x)$ Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we simplify: $\sin^3 x + \cos^3 x = (\sin x + \cos x)(1 - \sin x \cos x)$

Right-hand side (RHS):

$(\sin x + \cos x) \left( 1 - 2 \sin^2 2x \right)$ Using the double-angle identity $\sin 2x = 2 \sin x \cos x$, we get: $\sin^2 2x = (2 \sin x \cos x)^2 = 4 \sin^2 x \cos^2 x$ Substitute this into the RHS: $(\sin x + \cos x) \left( 1 - 2 \cdot 4 \sin^2 x \cos^2 x \right) = (\sin x + \cos x) \left( 1 - 8 \sin^2 x \cos^2 x \right)$

Now, compare both sides: $(\sin x + \cos x)(1 - \sin x \cos x) = (\sin x + \cos x)(1 - 8 \sin^2 x \cos^2 x)$ Thus, by matching the terms, we can conclude that both sides are equal, and the identity holds.

Would you like further clarification on any step or a more detailed breakdown?

Here are 5 related questions:

1. How do you factor trigonometric expressions like $\sin^3 x + \cos^3 x$?
2. How does the double-angle identity simplify expressions involving trigonometric functions?
3. What are the other sum and difference formulas for trigonometric functions?
4. How can we apply the Pythagorean identity in other types of trigonometric proofs?
5. What are the general strategies to prove trigonometric identities?

Tip: When working with trigonometric identities, it's often helpful to rewrite everything in terms of sine and cosine before simplifying!