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

I to be wondering what space other means to prove the $x^3-y^3 = (x-y)(x^2+xy+y^2)$


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Let $omega$ be a facility 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 as soon as $x in y,omega y,omega^2y $ and the degrees are right. Since $1 + omega + omega^2 = 0$ we have $omega + omega^2 = -1.$We additionally have $omega omega^2 = 1$, so we have $(x - omega y)(x-omega^2y) = x^2+xy + y^2.$


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They agree in ~ $,x = 0,pm y,$ so their distinction is a quadratic in $,x,$ v $3$ roots, for this reason zero.

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First, notice that for every $u$,eginalign*(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 multiply through $x^3$.


Well there space two ways that pertained to mind.

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

Another method would it is in to write:

$$left(fracxy ight)^3 - 1$$

Now we wish to find the zeros the this polynomial. These correspond come $fracxy = 1$, $fracxy = e^ifrac2pi3$ and also $fracxy = e^ifrac4pi3$.

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Then we can element the polynomial as:

$$left(fracxy ight)^3 - 1 = left( fracxy - 1 ight) left(fracxy - e^ifrac2pi3 ight) left( fracxy - e^ifrac4pi3 ight)$$

If us multiply the last two determinants together us find:

$$left(fracxy ight)^3 - 1 = left( fracxy - 1 ight) left(fracx^2y^2 - fracxy left(e^i frac2pi3 + e^i frac4 pi3 ight) + 1 ight)$$$$=left( fracxy - 1 ight) left(fracx^2y^2 - fracxy left( 2 cos(2pi/3) ight) + 1 ight) = left( fracxy - 1 ight) left(fracx^2y^2 + fracxy + 1 ight).$$

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