## anonymous one year ago 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

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