How do you factor ${sin}^{3} X - {cos}^{3} X$ $Sin^3X - Cos^3X$ ?

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Answer 1 · 2017-04-21T14:45:06

The answer is $= \frac{1}{2} (sin x - cos x) (2 + sin (2 x))$ $=1/2(sinx-cosx)(2+sin(2x))$

We apply

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $a^3-b^3=(a-b)(a^2+ab+b^2)$

Here,

$a = sin x$ $a=sinx$

and

$b = cos x$ $b=cosx$

So,

${sin}^{3} x - {cos}^{3} x = (sin x - cos x) ({sin}^{2} x + sin x cos x + {cos}^{2} x)$ $sin^3x-cos^3x=(sinx-cosx)(sin^2x+sinxcosx+cos^2x)$

But,

${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$

Therefore,

${sin}^{3} x - {cos}^{3} x = (sin x - cos x) (1 + sin x cos x)$ $sin^3x-cos^3x=(sinx-cosx)(1+sinxcosx)$

$= (sin x - cos x) (1 + \frac{sin 2 x}{2})$ $=(sinx-cosx)(1+(sin2x)/2)$

$= \frac{1}{2} (sin x - cos x) (2 + sin (2 x))$ $=1/2(sinx-cosx)(2+sin(2x))$

How do you factor Sin^3X - Cos^3Xsin3X−cos3XSin^3X - Cos^3X?