$x^3-y^3 = (x-y)x^2 + (x-y)(xy) + (x-y)y^2$ \implies x^3 - x^2y + x^2y -xy^2 + xy^2 - y^3\implies x^3 - y^3 I 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. You are watching: X^3-y^3 factor 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.