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

I to be wondering what room other methods to prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$


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Let $\omega$ be a complicated cube source of unity. Climate $x^3 - y^3 = (x-y)(x- \omega y)(x-\omega^2y)$ due to the fact that both sides vanish once $x \in \y,\omega y,\omega^2y \$ and also the levels are right. Since $1 + \omega + \omega^2 = 0$ we have $\omega + \omega^2 = -1.$We additionally have $\omega \omega^2 = 1$, for this reason we have actually $(x - \omega y)(x-\omega^2y) = x^2+xy + y^2.$


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They agree in ~ $\,x = 0,\pm y\,$ therefore their difference is a quadratic in $\,x\,$ v $3$ roots, hence zero.

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First, notification that for every $u$,\beginalign*(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.\endalign*Now, take $u = \fracyx$ and multiply through $x^3$.


Well there room two methods that concerned mind.

It is clear that when $x=y$ we have actually $x^3-y^3=0$. Then use long division to division $x^3-y^3$ through $x-y$ and the an outcome will it is in the equation ~ above the right.

Another method would be to write:

$$\left(\fracxy\right)^3 - 1$$

Now we wish to find the zeros the this polynomial. These correspond come $\fracxy = 1$, $\fracxy = e^i\frac2\pi3$ and $\fracxy = e^i\frac4\pi3$.

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Then us can aspect the polynomial as:

$$\left(\fracxy\right)^3 - 1 = \left( \fracxy - 1 \right) \left(\fracxy - e^i\frac2\pi3\right) \left( \fracxy - e^i\frac4\pi3 \right)$$

If we multiply the last two components together us find:

$$\left(\fracxy\right)^3 - 1 = \left( \fracxy - 1 \right) \left(\fracx^2y^2 - \fracxy \left(e^i \frac2\pi3 + e^i \frac4 \pi3\right) + 1 \right)$$$$=\left( \fracxy - 1 \right) \left(\fracx^2y^2 - \fracxy \left( 2 \cos(2\pi/3) \right) + 1 \right) = \left( \fracxy - 1 \right) \left(\fracx^2y^2 + \fracxy + 1 \right).$$

Thus $$\left(\fracxy\right)^3 - 1 = \left( \fracxy - 1 \right) \left(\fracx^2y^2 + \fracxy + 1 \right).$$ multiplying by $y^3$ ~ above both sides provides the result.