Solve for x cube root of 4x^2-4x+1- cube root of x=0

$\sqrt[3]{4 x^{2} - 4 x + 1} - \sqrt[3]{x} = 0$

Factor using the perfect square rule.

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

$\sqrt[3]{{(2 x)}^{2} - 4 x + 1} - \sqrt[3]{x} = 0$

Rewrite $1$ as $1^{2}$ .

$\sqrt[3]{{(2 x)}^{2} - 4 x + 1^{2}} - \sqrt[3]{x} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot (2 x) \cdot 1$

Rewrite the polynomial.

$\sqrt[3]{{(2 x)}^{2} - 2 \cdot (2 x) \cdot 1 + 1^{2}} - \sqrt[3]{x} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 2 x$ and $b = 1$ .

$\sqrt[3]{{(2 x - 1)}^{2}} - \sqrt[3]{x} = 0$

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

$x = \frac{1}{4}, 1$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{4}, 1$

Decimal Form:

$x = 0.25, 1$