Answer to Question #273383 in Calculus for 123

Question

Show that the transformation x = z −$\frac{b}{3a}$ converts the cubic equation ax³ + bx² + cx + d= 0 into one of the form z³ + 3Hz + G = 0, i.e. the x² term goes away.

Answer 1

$$x=z-\dfrac{b}{3a}$$

$$x^2=z^2-\dfrac{2bz}{3a}+\dfrac{b^2}{9a^2}$$

$$x^3=z^3-\dfrac{bz^2}{a}+\dfrac{b^2z}{3a^2}-\dfrac{b^3}{27a^3}$$

Substitute

$$az^3-bz^2+\dfrac{b^2z}{3a}-\dfrac{b^3}{27a^2}$$

$$+bz^2-\dfrac{2b^2z}{3a}+\dfrac{b^3}{9a^2}$$

$$cz-\dfrac{bc}{3a}+d=0$$

$$az^3+(-\dfrac{b^2}{3a}+c)z+(\dfrac{2b^3}{9a^2}-\dfrac{bc}{3a}+d)=0$$

$$z^3+3(-\dfrac{b^2}{3a^2}+\dfrac{c}{3a})z+(\dfrac{2b^3}{9a^3}-\dfrac{bc}{3a^2}+\dfrac{d}{a})=0$$

$$H=-\dfrac{b^2}{3a^2}+\dfrac{c}{3a}$$

$$G=\dfrac{2b^3}{9a^3}-\dfrac{bc}{3a^2}+\dfrac{d}{a}$$

$$z^3+3Hz+G=0$$