Write as a Function of x x^2+(y- cube root of x^2)^2=1

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

Step 1

Subtract $x^{2}$ from both sides of the equation.

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

Step 2

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$y - \sqrt[3]{x^{2}} = \pm \sqrt{1 - x^{2}}$

Step 3

Simplify $\pm \sqrt{1 - x^{2}}$ .

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

$y - \sqrt[3]{x^{2}} = \pm \sqrt{1^{2} - x^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = x$ .

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

Step 4

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

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

Add $\sqrt[3]{x^{2}}$ to both sides of the equation.

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

Next, use the negative value of the $\pm$ to find the second solution.

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

Add $\sqrt[3]{x^{2}}$ to both sides of the equation.

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

The complete solution is the result of both the positive and negative portions of the solution.

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

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

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

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

Step 5

To rewrite as a function of $x$ , write the equation so that $y$ is by itself on one side of the equal sign and an expression involving only $x$ is on the other side.

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

Write as a Function of x x^2+(y- cube root of x^2)^2=1

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