Question

How do you factor ${\sin }^{3}x-{\cos }^{3}x$?

Answer 1

Factorize using the common factor method

The given expression is ${\sin }^{3}x-{\cos }^{3}x$.

It can be factorized as follows,

${\sin }^{3}x-{\cos }^{3}x={\left(\sin x\right)}^3-{\left(\cos x\right)}^3$

$\Rightarrow {\sin }^{3}x-{\cos }^{3}x=\left(\sin x-\cos x\right)\left[{\left(\sin x\right)}^2+\sin x\times \cos x+{\left(\cos x\right)}^2\right]$ $[\because a^3-b^3=\left(a-b\right)\left(a^2+\mathrm{ab}+b^2\right)]$

$\Rightarrow {\sin }^{3}x-{\cos }^{3}x=\left(\sin x-\cos x\right)\left[{\sin }^{2}x+\sin x\cos x+{\cos }^{2}x\right]$

$\Rightarrow {\sin }^{3}x-{\cos }^{3}x=\left(\sin x-\cos x\right)\left(1+\sin x\cos x\right)$ $[\because {\sin }^{2}x+{\cos }^{2}x=1]$

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

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

$\Rightarrow {\sin }^{3}x-{\cos }^{3}x=\frac{1}{2}\left(\sin x-\cos x\right)\left(2+\sin 2x\right)$ $\left[\because \sin 2x=2\sin x\cos x\right]$

Hence, the factors of ${\sin }^{3}x-{\cos }^{3}x$ are $\frac{1}{2}\left(\sin x-\cos x\right)\left(2+\sin 2x\right)$.