# Solving Cubic Equations

iSolve : For A,B,C reals

$x^3 + Ax^2 + Bx + C =0$

Method :

Let $y = x-k \implies x=y+k$

$\implies (y+k)^3 + A(y+k)^2 + B(y+k) + C =0$

$\implies y^3 + (3k+A)y^2 + (3k^2+2Ak+B)y + (k^3+Ak^2+Bk +C) =0$

Choose , $3k+A=0 \implies k=\frac{-A}{3}$

$\implies y^3 + (B-\frac{A^2}{3})y + \frac{2A^3}{27}-\frac{AB}{3}+C = 0$

rewrite this as –

$\implies y^3 +py + q =0$

which again is re-written as –

$\implies y^3+a^3+b^3-3aby=0$

Where ,

$p = -3ab , q= a^3 + b^3$

$\implies p = -27a^3b^3 , q= a^3 + b^3$

Therefore, $a^3$ and $b^3$ are the roots of the equation –

$z^2 - qz - \frac{p^3}{27}=0$

Therefore ,

$a = \sqrt[3]{\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{4p^3}{27}}}$

$b = \sqrt[3]{\frac{q}{2}-\sqrt{\frac{q^2}{4}-\frac{4p^3}{27}}}$

Now we have that –

$y^3 +a^3+b^3-3aby = (a+b+y)(y^2-(a+b)y +a^2-ab+b^2) = 0$

Solving the quadratic plus the trivial solution gives us –

$y = -(a+b) , \frac{a+b}{2}+\frac{\sqrt{3}}{2}(1-b)i , \frac{a+b}{2}-\frac{\sqrt{3}}{2}(a-b)i$

now substituting the value of y in $x = y + k$ we get the three very simple roos (:D)

I know.. the calculations are extremely annoying…… perhaps that is why this is used very less…