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

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

Step 1

Simplify ${(x + 1)}^{2}$ .

Rewrite.

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

Rewrite ${(x + 1)}^{2}$ as $(x + 1) (x + 1)$ .

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

Expand $(x + 1) (x + 1)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

$- 2 y + \sqrt[3]{x} = x \cdot x + x \cdot 1 + 1 x + 1 \cdot 1$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

Multiply $x$ by $1$ .

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

Multiply $x$ by $1$ .

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

Multiply $1$ by $1$ .

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

Add $x$ and $x$ .

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

Step 2

Subtract $\sqrt[3]{x}$ from both sides of the equation.

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

Step 3

Divide each term in $- 2 y = x^{2} + 2 x + 1 - \sqrt[3]{x}$ by $- 2$ and simplify.

Divide each term in $- 2 y = x^{2} + 2 x + 1 - \sqrt[3]{x}$ by $- 2$ .

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

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

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

Cancel the common factor of $2$ and $- 2$ .

Factor $2$ out of $2 x$ .

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

Move the negative one from the denominator of $\frac{x}{- 1}$ .

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

Rewrite $- 1 \cdot x$ as $- x$ .

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

Move the negative in front of the fraction.

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

Dividing two negative values results in a positive value.

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

Step 4

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.

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

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

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