# How to prove $x^3-y^3 = (x-y)(x^2+xy+y^2)$ without expand the right side?

I can prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$ by expanding the right side.

1. $x^3-y^3 = (x-y)x^2 + (x-y)(xy) + (x-y)y^2$
2. $\implies x^3 - x^2y + x^2y -xy^2 + xy^2 - y^3$
3. $\implies x^3 - y^3$

I was wondering what are other ways to prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$

Let $\omega$ be a complex cube root of unity. Then $x^{3} - y^{3} = (x-y)(x- \omega y)(x-\omega^{2}y)$ since both sides vanish when $x \in \{y,\omega y,\omega^{2}y \}$ and the degrees are right. Since $1 + \omega + \omega^{2} = 0$ we have $\omega + \omega^{2} = -1.$ We also have $\omega \omega^{2} = 1$, so we have $(x - \omega y)(x-\omega^{2}y) = x^{2}+xy + y^{2}.$

Divide $x^3-y^3$ by $x-y$ as polynomials

First, notice that for all $u$, \begin{align*}(1-u)(1 + u + u^2) &= 1 + u + u^2 - (u - u^2 - u^3)\\ &= 1 + (u-u) + (u^2-u^2) - u^3\\ &= 1 - u^3.\end{align*} Now, take $u = \frac{y}{x}$ and multiply by $x^3$.

They agree at $\,x = 0,\pm y\,$ so their difference is a quadratic in $\,x\,$ with $3$ roots, hence zero.

• Here the coefficient ring is the integral domain $\,\Bbb Z[y]\,$ where $\,0,y,-y\,$ are distinct. Recall that a nonzero polynomial over a domain has no more roots than its degree. – Bill Dubuque Mar 2 '15 at 18:32

Since $x=y$ is clearly a root, one may divide $x^3-y^3$ by $x-y$ directly.

Well there are two ways that come to mind.

It is clear that when $x=y$ we have $x^3-y^3=0$. Then use long division to divide $x^3-y^3$ by $x-y$ and the result will be the equation on the right.

Another way would be to write:

$$\left(\frac{x}{y}\right)^3 - 1$$

Now we wish to find the zeros of this polynomial. These correspond to $\frac{x}{y} = 1$, $\frac{x}{y} = e^{i\frac{2\pi}{3}}$ and $\frac{x}{y} = e^{i\frac{4\pi}{3}}$.

Then we can factor the polynomial as:

$$\left(\frac{x}{y}\right)^3 - 1 = \left( \frac{x}{y} - 1 \right) \left(\frac{x}{y} - e^{i\frac{2\pi}{3}}\right) \left( \frac{x}{y} - e^{i\frac{4\pi}{3}} \right)$$

If we multiply the last two factors together we find:

$$\left(\frac{x}{y}\right)^3 - 1 = \left( \frac{x}{y} - 1 \right) \left(\frac{x^2}{y^2} - \frac{x}{y} \left(e^{i \frac{2\pi}{3}} + e^{i \frac{4 \pi}{3}}\right) + 1 \right)$$ $$=\left( \frac{x}{y} - 1 \right) \left(\frac{x^2}{y^2} - \frac{x}{y} \left( 2 \cos(2\pi/3) \right) + 1 \right) = \left( \frac{x}{y} - 1 \right) \left(\frac{x^2}{y^2} + \frac{x}{y} + 1 \right).$$

Thus $$\left(\frac{x}{y}\right)^3 - 1 = \left( \frac{x}{y} - 1 \right) \left(\frac{x^2}{y^2} + \frac{x}{y} + 1 \right).$$ Multiplying by $y^3$ on both sides gives the result.

Another way is as follows: $$(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3 = (x^3 - y^3) -3xy(x - y) \quad \Rightarrow$$ $$x^3 - y^3 = (x - y)^3 + 3xy(x - y) = (x - y)[(x - y)^2 + 3xy] \quad \Rightarrow$$ $$x^3 - y^3 = (x - y)(x^2 + xy + y^2)$$

$x^3-y^3=x^2(x-y)+yx^2-y^3=x^2(x-y)+yx(x-y)+xy^2-y^3=...$ (Simly insert $x^2y-yx^2$, then insert $y^2x-xy^2$,...)

You can rearrange it into two possible ways.

Show, by left side, that $$\frac{x^3-y^3}{x-y} = x^2+xy+y^2,$$ or $$\frac{x^3-y^3}{x^2+xy+y^2} = x-y.$$

You could do the enclidian division of $X^3 - Y^3$ by $X-Y$ in the ring $A[X]$, where $A = \mathbf{Z}[Y]$, has $X-Y$ has unit leading coefficient.

You can use "Long Division of Polynomials"

$\frac {1-(\frac y x)^3}{1-(\frac y x)}=1+\frac y x+(\frac y x)^2$

Multiply both sides by $x^3$

$\frac {x^3-y^3}{1-(\frac y x)}=x(x^2+xy+y^2)$

Then you have

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

• What's the point of using $\frac yx$ in the first place? Why can't you just perform your long division on $1-x^3$? – BigbearZzz Aug 8 '16 at 8:21