## Trigonometry (11th Edition) Clone

$$\sin^3x-\cos^3x=(\sin x-\cos x)(1+\sin x\cos x)$$
$$A=\sin^3x-\cos^3x$$ Now it is crucial here not to forget that $$a^3-b^3=(a-b)(a^2+ab+b^2)$$ which means $$A=(\sin x-\cos x)(\sin^2 x+\sin x\cos x+\cos^2 x)$$ $$A=(\sin x-\cos x)[(\sin^2 x+\cos^2 x)+\sin x\cos x]$$ - From Pythagorean Identity: $$\sin^2 x+\cos^2x=1$$ So, $A$ would be $$A=(\sin x-\cos x)(1+\sin x\cos x)$$