# How do you find the real zeros of y=3/4(x-2/3)^3-16/9?

Mar 7, 2018

2

#### Explanation:

You rewrite $\frac{3}{4} {\left(x - \frac{2}{3}\right)}^{3} - \frac{16}{9} = 0$ in the form :

$\frac{3}{4} {\left(x - \frac{2}{3}\right)}^{3} = \frac{16}{9}$

so that

${\left(x - \frac{2}{3}\right)}^{3} = \frac{16}{9} \times \frac{4}{3} = {\left(\frac{4}{3}\right)}^{3}$

The only real solution to ${x}^{3} = {a}^{3}$ is $x = a$ (the other two are $x = a \omega$ and $x = a {\omega}^{2}$, where $\omega = \exp \left(\frac{2 \pi}{3} i\right)$ and ${\omega}^{2}$ are the complex cube roots of unity), so

$x - \frac{2}{3} = \frac{4}{3}$
and thus
$x = \frac{2}{3} + \frac{4}{3} = 2$