Answer
See the explanation below.
Work Step by Step
Let us consider the left-hand side of the given expression:
${{\sin }^{3}}x+{{\cos }^{3}}x$
So, according to the algebraic formula, which is, ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$, ${{\sin }^{3}}x+{{\cos }^{3}}x$ can be written as:
${{\sin }^{3}}x+{{\cos }^{3}}x=\left( \sin x+\cos x \right)\left( {{\sin }^{2}}x+{{\cos }^{2}}x-\sin x\cos x \right)$
Now, apply the Pythagorean identity, that is ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ to further simplify the expression as given below:
${{\sin }^{3}}x+{{\cos }^{3}}x=\left( \sin x+\cos x \right)\left( 1-\sin x\cos x \right)\text{ }$
Then, multiply and divide the expression $\sin x\cos x$ by $2$:
$\begin{align}
& {{\sin }^{3}}x+{{\cos }^{3}}x=\left( \sin x+\cos x \right)\left( 1-\frac{2}{2}\sin x\cos x \right) \\
& =\left( \sin x+\cos x \right)\left( 1-\frac{2\sin x\cos x}{2} \right)\text{ }
\end{align}$
One of the double angle formulas is $\sin 2x=2\sin x\cos x$.
Therefore, applying this formula, further simplification of the expression can be done as:
${{\sin }^{3}}x+{{\cos }^{3}}x=\left( \sin x+\cos x \right)\left( 1-\frac{\sin 2x}{2} \right)$
Thus, the left-hand side of the expression is equal to the right-hand side, which is ${{\sin }^{3}}x+{{\cos }^{3}}x=\left( \sin x+\cos x \right)\left( 1-\frac{\sin 2x}{2} \right)$.
