How do you prove sin^3 x - cos^3 x = (1 + sin x cos x )( sin x cos x )?

Jun 18, 2015

This is a mistaken identity. Substitute $x = 0$ - left part equals to ${\cos}^{3} \left(0\right) = - 1$ while the right part equals to $0$ since $\sin \left(0\right) = 0$.
The right identity is
${\sin}^{3} x - {\cos}^{3} x = \left(1 + \sin x \cdot \cos x\right) \left(\sin x - \cos x\right)$

Explanation:

Recall the trivial algebraic formula
${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$
It can be verified by direct multiplication in the right part.

Applying this to our problem for $a = \sin x$ and $b = \cos x$, we obtain
${\sin}^{3} x - {\cos}^{3} x = \left(\sin x - \cos x\right) \left({\sin}^{2} x + \sin x \cdot \cos x + {\cos}^{2} x\right)$

Since ${\sin}^{2} x + {\cos}^{2} x = 1$, the expression in the second pair of parenthesis in the right part of this identity can be simplified:
${\sin}^{2} x + \sin x \cdot \cos x + {\cos}^{2} x = 1 + \sin x \cdot \cos x$.
This completes the proof.