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…

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