The equation whose roots are the squares of the roots of $x^{3}+ax+b=0$, is
- $x^{3}+2ax^{2}-a^{2}x-b^{2}=0$
- $x^{3}+2ax^{2}+a^{2}x-b^{2}=0$
- $x^{6}+ax^{2}+6=0$
- $x^{6}-ax^{2}+6=0$
Let y = x^2, so x = sqrt(y). Substituting into x^3 + ax + b = 0 gives y*sqrt(y) + a*sqrt(y) + b = 0, or sqrt(y)(y + a) = -b. Squaring both sides: y(y + a)^2 = b^2, which is y(y^2 + 2ay + a^2) = b^2, or y^3 + 2ay^2 + a^2y - b^2 = 0.
Let the original roots be alpha, beta, and gamma, meaning the sum of the roots is 0, the sum of products taken two at a time is a, and the product is minus b. We need a new cubic equation whose roots are alpha squared, beta squared, and gamma squared, so we find the new symmetric sums. The new sum of roots is alpha squared plus beta squared plus gamma squared, which equals (alpha plus beta plus gamma) squared minus 2 times the sum of products, yielding minus 2a. The new sum of products taken two at a time is (alpha beta) squared plus (beta gamma) squared plus (gamma alpha) squared, which equals a squared. The new product of the roots is (alpha beta gamma) squared, which is b squared. Substituting these values into a standard cubic x cubed minus S1 x squared plus S2 x minus S3 equals 0, we get x cubed plus 2a x squared plus a squared x minus b squared equals 0.