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Prove this $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$

My attempt

Proof - by using [axiomdistributive] and [axiommulcommutative]: $$\begin{split} &(x+y)(x^2 - xy + y^2)\\ &= (x+y)x^2 - (x+y)xy + (x+y)y^2\\ &= (x^3+x^2y) - (x^2y+xy^2) + (xy^2+y^3)\\ &= x^3 + x^2y - x^2y - xy^2 + xy^2 + y^3\\ &= x^3 + y^3\\ \end{split}$$

Q.E.D.

Question:

Spivak says there is an easy proof that, if I use this other theorem:

$$ x^3 - y^3 = (x-y)(x^2 + xy + y^2) $$

then, I will also allow me to find out $x^n+y^n$ whenever $n$ is odd.

How to do this? I fail to see how.

YuiTo Cheng
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caveman
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4 Answers4

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In this case of $n=3$ your result follows from this other theorem by considering $x^3+y^3=x^3-(-y)^3$. The pattern with odd $n$ is that $-y^n=(-y)^n$.

Dave
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Maybe not easier, but given $$\frac{1-z^n}{1-z}=\sum_{i=0}^{n-1}z^i\\1-z^n=(1-z)\sum_{i=0}^{n-1}z^i\\$$ let $z=\frac{y}{x}$, then $$1-\left(\frac{y}{x}\right)^n=\left(1-\frac{y}{x}\right)\sum_{i=0}^{n-1}y^ix^{-i}$$ multiplying by $x^n$ yields $$x^n-y^n=(x-y)\sum_{i=0}^{n-1}y^ix^{n-1-i}$$

cansomeonehelpmeout
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$$x^3+y^3 = x^3+\color{red}{x^2y-x^2y}+y^3 $$ $$ = x^2(x+y)-y(x^2-y^2)$$ $$ = x^2(x+y)-y(x-y)(x+y)$$ $$ = (x+y)(x^2-y(x-y))$$ $$ = (x+y)(x^2-xy+y^2)$$

We can proceede similary for arbitrary odd $n$.

nonuser
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Just let $k=-y$ to obtain $$x^3-y^3=(x-y)(x^2+xy+y^2)\iff x^3+k^3=(x+k)(x^2-xk+k^2)$$ Note that the same works whenever $n$ is odd. Given $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\dots + xy^{n-2}+y^{n-1})$$ Let $k=-y$ and ...

See how it works? Try to explain why it doesn't work whenever $n$ is even!

Dr. Mathva
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