# How do you solve 3x ^(2/3) + x^(1/3) - 2 = 0?

$x = \frac{8}{27}$ and $x = - 1$

#### Explanation:

From the given equation
$3 {x}^{\frac{2}{3}} + {x}^{\frac{1}{3}} - 2 = 0$

The solution goes like this

Let $w = {x}^{\frac{1}{3}}$

$3 {x}^{\frac{2}{3}} + {x}^{\frac{1}{3}} - 2 = 0$

we can write this equation this way

$3 {\left({x}^{\frac{1}{3}}\right)}^{2} + {\left({x}^{\frac{1}{3}}\right)}^{1} - 2 = 0$

also

$3 {\left(w\right)}^{2} + 1 \cdot {\left(w\right)}^{1} - 2 = 0$

also let $a = 3$ and $b = 1$ and $c = - 2$

$w = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$w = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \cdot 3 \cdot \left(- 2\right)}}{2 \cdot 3}$

$w = \frac{- 1 \pm \sqrt{1 + 24}}{6}$

$w = \frac{- 1 \pm \sqrt{25}}{6}$

$w = \frac{- 1 \pm 5}{6}$

we have two values for w:

${w}_{1} = \frac{- 1 + 5}{6} = \frac{4}{6} = \frac{2}{3}$ and ${w}_{2} = \frac{- 1 - 5}{6} = - \frac{6}{6} = - 1$

But $w = {x}^{\frac{1}{3}}$

and $x = {w}^{3}$

Using $w = \frac{2}{3}$

$x = {w}^{3} = {\left(\frac{2}{3}\right)}^{3} = \frac{8}{27}$
$x = \frac{8}{27}$

Using $w = - 1$

$x = {w}^{3} = {\left(- 1\right)}^{3} = - 1$

$x = - 1$

check at $x = \frac{8}{27}$
$3 {x}^{\frac{2}{3}} + {x}^{\frac{1}{3}} - 2 = 0$
$3 {\left(\frac{8}{27}\right)}^{\frac{2}{3}} + {\left(\frac{8}{27}\right)}^{\frac{1}{3}} - 2 = 0$
$3 \left(\frac{4}{9}\right) + \frac{2}{3} - 2 = 0$
$\frac{4}{3} + \frac{2}{3} - 2 = 0$
$\frac{6}{3} - 2 = 0$
$2 - 2 = 0$
$0 = 0$
$x = \frac{8}{27}$ is a root

check at $x = - 1$
$3 {x}^{\frac{2}{3}} + {x}^{\frac{1}{3}} - 2 = 0$
$3 {\left(- 1\right)}^{\frac{2}{3}} + {\left(- 1\right)}^{\frac{1}{3}} - 2 = 0$
$3 \left(1\right) - 1 - 2 = 0$
$3 - 3 = 0$
$0 = 0$

$x = - 1$ is a root

God bless....I hope the explanation is useful.