Precalculus (6th Edition) Blitzer

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