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Prove the identity: $$ \cos^3x+\sin^3x=(\cos x+\sin x)\cdot(1-\sin x\cdot\cos x) $$ I was able to change both sides to $\cos x-\cos x\cdot\sin^{2}x+\sin x-\sin x\cdot\cos^{2}x$, which is kind of long. Is there a shorter way, such as factoring the left side? If so, how can I do it?

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HINT: Have you seen an identity involving $a^3 + b^3$? – Srivatsan Dec 7 '11 at 0:24
Maybe you have seen a factorization of $x^3-y^3$? – André Nicolas Dec 7 '11 at 0:26
Then the phrase you want to search for is "sum of cubes" (or "difference of cubes"). – Blue Dec 7 '11 at 0:30
So you don't know $x^3+y^3=(x+y)(x^2-xy+y^2)$? – J. M. Dec 7 '11 at 0:31
Alex Yan: I'm not sure exactly if this is what André intended, but $x^3+y^3$ also has the form $x^3-z^3$ with $z=-y$; if you know one factorization you get the other. – Jonas Meyer Dec 7 '11 at 0:35
up vote 7 down vote accepted

Observe that $$ \begin{align*} \cos^3x+\sin^3x &= (\cos x+\sin x)(\cos^2x-\sin x \cos x +\sin^2x) \\ &= (\cos x+\sin x)(1-\sin x \cos x ) \end{align*}$$ noting that $\cos^2x+ \sin^2x=1$.

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