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))#