How do you make Z the subject of the formula in Z3−3YZ−3X=0?

Question

How do you make Z the subject of the formula in  Z3−3YZ−3X=0?

Jillian Edgerton · Accepted Answer

This has a depressed cubic shape. The secret to resolving such an equation is as follows:
$Z^{3} - 3 Y Z = 3 X$
Let's begin with the cubed binomial form.
${(p + q)}^{3} = p^{3} + 3 p^{2} q + 3 p q^{2} + q^{3}$
If we subtract both sides by the middle 2 terms of the expansion, we get:
${(p + q)}^{3} - 3 p^{2} q - 3 p q^{2} = p^{3} + q^{3}$
${(p + q)}^{3} - 3 p^{2} q - 3 p q^{2} = p^{3} + q^{3}$
We can make some substitutions:
$p + q = Z$
$3 p q = 3 Y$
$p^{3} + q^{3} = 3 X$
That makes this match the equation we are trying to solve.
From the second substitution equation, we get:
$p = \frac{Y}{q}$
Plugging that into the third substitution equation, we get:
$\frac{Y^{3}}{q^{3}} + q^{3} = 3 X$
That becomes a quadratic that can be solved for $q^{3}$ .
$q^{3} = \frac{3 X \pm \sqrt{9 X^{2} - 4 (1) (Y^{3})}}{2 (1)}$
$q^{3} = \frac{3 X}{2} \pm \sqrt{\frac{9 X^{2}}{4} - Y^{3}}$
$q = \sqrt[3]{\frac{3 X}{2} \pm \sqrt{\frac{9 X^{2}}{4} - Y^{3}}}$
One of those solutions will be q, the other will be p.
Plugging those into the first of the 3 substitution equations, we get:
$Z = \sqrt[3]{\frac{3 X}{2} + \sqrt{\frac{9 X^{2}}{4} - Y^{3}}} + \sqrt[3]{\frac{3 X}{2} - \sqrt{\frac{9 X^{2}}{4} - Y^{3}}}$
Having that solution, you can use polynomial division to get a quadratic to determine the other 2 roots from the original equation.

How do you make Z the subject of the formula

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