# How do you find all the real and complex roots of x^3 + 3x^2 + 3x + 3 = 0?

Feb 12, 2016

Recognise the resemblance to ${\left(x + 1\right)}^{3}$ to simplify the cubic and solve.

#### Explanation:

We use the sum of cubes identity below, which can be written in the form:

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

One way to solve the cubic equation is to notice that it is rather similar to the binomial expansion of ${\left(x + 1\right)}^{3}$ ...

$0 = {x}^{3} + 3 {x}^{2} + 3 x + 3$

$= {x}^{3} + 3 {x}^{2} + 3 x + 1 + 2$

$= {\left(x + 1\right)}^{3} + 2$

$= {\left(x + 1\right)}^{3} + {\sqrt[3]{2}}^{3}$

$= \left(\left(x + 1\right) + \sqrt[3]{2}\right) \left({\left(x + 1\right)}^{2} - \sqrt[3]{2} \left(x + 1\right) + {\sqrt[3]{2}}^{2}\right)$

$= \left(x + \left(1 + \sqrt[3]{2}\right)\right) \left({x}^{2} + \left(2 - \sqrt[3]{2}\right) x + \left(1 - \sqrt[3]{2} + {\sqrt[3]{2}}^{2}\right)\right)$

So the Real root is $x = - 1 - \sqrt[3]{2}$ and the Complex roots are given by the quadratic formula:

$x = \frac{\left(\sqrt[3]{2} - 2\right) \pm \sqrt{{\left(2 - \sqrt[3]{2}\right)}^{2} - 4 \left(1 - \sqrt[3]{2} + {\sqrt[3]{2}}^{2}\right)}}{2}$

$= \frac{\left(\sqrt[3]{2} - 2\right) \pm \sqrt{4 - 2 \sqrt[3]{2} + {\sqrt[3]{2}}^{2} - 4 + 4 \sqrt[3]{2} - 4 {\sqrt[3]{2}}^{2}}}{2}$

$= \frac{\left(\sqrt[3]{2} - 2\right) \pm \sqrt{2 \sqrt[3]{2} - 3 {\sqrt[3]{2}}^{2}}}{2}$

$= \frac{\sqrt[3]{2} - 2}{2} \pm \frac{\sqrt{3 {\sqrt[3]{2}}^{2} - 2 \sqrt[3]{2}}}{2} i$