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

Let $\omega$ it is in a facility cube root of unity. Then $x^3 - y^3 = (x-y)(x- \omega y)(x-\omega^2y)$ because both political parties vanish as soon as $x \in \y,\omega y,\omega^2y \$ and the degrees are right. Since $1 + \omega + \omega^2 = 0$ we have actually $\omega + \omega^2 = -1.$We also have $\omega \omega^2 = 1$, for this reason we have actually $(x - \omega y)(x-\omega^2y) = x^2+xy + y^2.$

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

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First, notice that for all $u$,\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 it $u = \fracyx$ and also multiply by $x^3$.

Well there space two methods that come to mind.

It is clear that as soon as $x=y$ we have $x^3-y^3=0$. Then use long division to division $x^3-y^3$ by $x-y$ and the an outcome will it is in the equation on the right.

Another means would be to write:

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

Now us wish to discover the zeros that this polynomial. This correspond to $\fracxy = 1$, $\fracxy = e^i\frac2\pi3$ and $\fracxy = e^i\frac4\pi3$.

See more: What Is The Lowest Common Multiple Of 10 And 15 And 18), Least Common Multiple Of 10 And 15

Then we can element 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 us multiply the last two determinants 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 $y^3$ top top both sides offers the result.