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

You are watching: X^3+y^3 factor 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.