If one root of the equation $ax^2+bx+c=0$ is a square of the other other, then $a(b-c) ^3=cX$, where $X$ is
- $a^3-b^3$
- $a^3+b^3$
- $(a-b) ^3$
- $(b-a) ^3$
If the roots are alpha and alpha^2, then alpha + alpha^2 = -b/a and alpha^3 = c/a. From the first, alpha(1 + alpha) = -b/a. Cubing both sides: alpha^3(1 + alpha)^3 = -b^3/a^3. Substituting c/a for alpha^3 gives (c/a)(1 + alpha)^3 = -b^3/a^3. This leads to the identity a(b-c)^3 = c(b-a)^3.
Let the roots be r and r squared, so their sum is r plus r squared equals negative b over a, and their product is r cubed equals c over a. From the sum equation, multiplying by a gives a r cubed plus a r squared equals negative b, so a r cubed plus b equals negative a r squared. Cubing both sides of the product equation gives r to the power of 9 equals c cubed over a cubed, and cubing the sum equation gives a cubed r cubed plus a cubed r squared cubed equals negative b cubed. This leads to the symmetric identity a times the quantity b minus c cubed equals c times the quantity b minus a cubed, so the value of X is the quantity b minus a cubed.