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So I'm reading Beardon's Algebra and Geometry, and in chapter on complex numbers, author gives the following method for solving cubic equation:

Suppose we want to solve cubic equation $p_1(z)=0$, where $p_1(z)=z^3+az^2+bz+c$. Now $p_1(z-\frac{a}{3})$ is a cubic polynomial with no term in $z^2$, so we can consider it and relabel the coefficients to find the zeros of $p$, where now


The advantage of it is that


We can solve this quadreatic to obtain $z^3$ and hence a value, say $h$ of $z$. As $h^3-\frac{b^3}{h^3}=c$, it follows that $p(h-b/h)=0$ and we found a solution of $p(z)$

Then I'm asked to solve equation $z^3-z^2+z-1=0$ using this method.

So basically I do the first "transformation" with $p_1(z-\frac{a}{3})$. Then consider it with $p(z)=z^3+3bz-c$. But what I get for z as solutions is nowhere near to what the real solutions are. So I was wondering if somebody could provide me with a solution using this method?

Or is it that solutions for $p(z)$ will not be the same as the ones required for the initial equation and I need to do something additional to that? Thanks!

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up vote 3 down vote accepted

Let $p(z) = z^3 - z^2 + z - 1$. Following the method you described, we do the change of variables $z = w + \frac{1}{3}$ to get $$p(z) = w^3 + w^2 + \frac{w}{3} + \frac{1}{27} - w^2 - \frac{2w}{3} - \frac{1}{9} + w + \frac{1}{3} - 1 = w^3 + \frac{2}{3} w - \frac{20}{27}.$$ Now doing another change of variables, $w = u - \frac{2}{9u}$, we have $$p(z) = u^3 - \frac{2}{3}u+\frac{4}{27u}-\frac{8}{729u^3} + \frac{2}{3}u - \frac{4}{27u} - \frac{20}{27} = u^3 - \frac{8}{729u^3} - \frac{20}{27}.$$ Solving $u^3 - \frac{8}{729u^3} - \frac{20}{27} = 0$, which is equivalent with $u^6 - \frac{20}{27}u^3 - \frac{8}{729} = 0$, we have $$u^3 = \frac{\frac{20}{27} \pm \sqrt{\frac{400}{729} + \frac{32}{729}}}{2} = \frac{10}{27} \pm \frac{\sqrt{108}}{27}.$$ Thus we have six solutions for $u$, namely $$u = \frac{\sqrt[3]{10 + 6\sqrt{3}}}{27}\cdot e^{\frac{2\pi}{3} k i} \quad\text{and}\quad\frac{\sqrt[3]{10 - 6\sqrt{3}}}{27} \cdot e^{\frac{2\pi}{3} k i}, \quad k=0,1,2.$$ By substituting these back to the formula $w = u - \frac{2}{9u}$ and simplifying we find that $w \in \{\frac{2}{3}, -\frac{1}{3} + i, -\frac{1}{3} - i\}$, so in particular $z \in \{1, i, -i\}$.

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thank you. forgot to substitute back basically :) – Sarunas May 1 '13 at 17:47

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