Solve for x 4( cube root of x-2)+1=9

$4 (\sqrt[3]{x - 2}) + 1 = 9$

Step 1

Move all terms not containing $4 \sqrt[3]{x - 2}$ to the right side of the equation.

Subtract $1$ from both sides of the equation.

$4 \sqrt[3]{x - 2} = 9 - 1$

Subtract $1$ from $9$ .

$4 \sqrt[3]{x - 2} = 8$

Step 2

To remove the radical on the left side of the equation, cube both sides of the equation.

${(4 \sqrt[3]{x - 2})}^{3} = 8^{3}$

Step 3

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x - 2}$ as ${(x - 2)}^{\frac{1}{3}}$ .

${(4 {(x - 2)}^{\frac{1}{3}})}^{3} = 8^{3}$

Simplify the left side.

Simplify ${(4 {(x - 2)}^{\frac{1}{3}})}^{3}$ .

Apply the product rule to $4 {(x - 2)}^{\frac{1}{3}}$ .

$4^{3} {({(x - 2)}^{\frac{1}{3}})}^{3} = 8^{3}$

Raise $4$ to the power of $3$ .

$64 {({(x - 2)}^{\frac{1}{3}})}^{3} = 8^{3}$

Multiply the exponents in ${({(x - 2)}^{\frac{1}{3}})}^{3}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$64 {(x - 2)}^{\frac{1}{3} \cdot 3} = 8^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$64 {(x - 2)}^{\frac{1}{3} \cdot 3} = 8^{3}$

Rewrite the expression.

$64 {(x - 2)}^{1} = 8^{3}$

Simplify.

$64 (x - 2) = 8^{3}$

Apply the distributive property.

$64 x + 64 \cdot - 2 = 8^{3}$

Multiply $64$ by $- 2$ .

$64 x - 128 = 8^{3}$

Simplify the right side.

Raise $8$ to the power of $3$ .

$64 x - 128 = 512$

Step 4

Solve for $x$ .

Move all terms not containing $x$ to the right side of the equation.

Add $128$ to both sides of the equation.

$64 x = 512 + 128$

Add $512$ and $128$ .

$64 x = 640$

Divide each term in $64 x = 640$ by $64$ and simplify.

Divide each term in $64 x = 640$ by $64$ .

$\frac{64 x}{64} = \frac{640}{64}$

Simplify the left side.

Cancel the common factor of $64$ .

Cancel the common factor.

$\frac{64 x}{64} = \frac{640}{64}$

Divide $x$ by $1$ .

$x = \frac{640}{64}$

Simplify the right side.

Divide $640$ by $64$ .

$x = 10$

Solve for x 4( cube root of x-2)+1=9

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