Historical Feature:
Show that the general cubic equation y^3+by^2+cy+d=0 can be transformed into an equation of the form x^3+px +q =0 by using the substitution y=x-b/3

Question

Historical Feature: 
Show that the general cubic equation y^3+by^2+cy+d=0 can be transformed into an equation of the form x^3+px +q =0 by using the substitution y=x-b/3

mathstudent55 · Answer

$\left(x - \dfrac{b}{3} \right)^3 + b\left(x - \dfrac{b}{3} \right)^2 + c\left(x - \dfrac{b}{3} \right) + d=0$

$x^3 - bx^2 + \dfrac{b^2x}{3}  - \dfrac{b^3x}{27} + bx^2 - \dfrac{2b^2x}{3} + \dfrac{b^3}{9} + cx - \dfrac{bc}{3} + d = 0$

$x^3 + \dfrac{b^2x}{3}  - \dfrac{b^3x}{27} - \dfrac{2b^2x}{3} + \dfrac{b^3}{9} + cx - \dfrac{bc}{3} + d=0$

$x^3 + \left(\dfrac{b^2}{3}  - \dfrac{b^3}{27} - \dfrac{2b^2}{3} + c\right)x + \dfrac{b^3}{9} - \dfrac{bc}{3} + d=0$

Where

$p = \dfrac{b^2}{3}  - \dfrac{b^3}{27} - \dfrac{2b^2}{3} + c$

and 

$q = \dfrac{b^3}{9} - \dfrac{bc}{3} + d$