Solve for x 1/2 cube root of x^3+x^2=x

$\frac{1}{2} \sqrt[3]{x^{3} + x^{2}} = x$

Step 1

Simplify $\frac{1}{2} \cdot \sqrt[3]{x^{3} + x^{2}}$ .

Factor $x^{2}$ out of $x^{3} + x^{2}$ .

Factor $x^{2}$ out of $x^{3}$ .

$\frac{1}{2} \cdot \sqrt[3]{x^{2} x + x^{2}} = x$

Multiply by $1$ .

$\frac{1}{2} \cdot \sqrt[3]{x^{2} x + x^{2} \cdot 1} = x$

Factor $x^{2}$ out of $x^{2} x + x^{2} \cdot 1$ .

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

Combine $\frac{1}{2}$ and $\sqrt[3]{x^{2} (x + 1)}$ .

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

Step 2

Graph each side of the equation. The solution is the x-value of the point of intersection.

$x = \frac{1}{7}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{7}$

Decimal Form:

$x = 0.\overline{142857}$

Step 4