How do you factor $Sin^3X - Cos^3X$ ?

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Answer 1 · 2017-04-21T14:45:06

The answer is $=1/2(sinx-cosx)(2+sin(2x))$

We apply

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

Here,

$a=sinx$

and

$b=cosx$

So,

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

But,

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

Therefore,

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

$=(sinx-cosx)(1+(sin2x)/2)$

$=1/2(sinx-cosx)(2+sin(2x))$

How do you factor Sin^3X - Cos^3XSin^3X - Cos^3X?