Solve for x 4 cube root of 2x+1=20

$4 \sqrt[3]{2 x + 1} = 20$

Step 1

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

${(4 \sqrt[3]{2 x + 1})}^{3} = 20^{3}$

Step 2

Simplify each side of the equation.

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

${(4 {(2 x + 1)}^{\frac{1}{3}})}^{3} = 20^{3}$

Simplify the left side.

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

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

$4^{3} {({(2 x + 1)}^{\frac{1}{3}})}^{3} = 20^{3}$

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

$64 {({(2 x + 1)}^{\frac{1}{3}})}^{3} = 20^{3}$

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

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

$64 {(2 x + 1)}^{\frac{1}{3} \cdot 3} = 20^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$64 {(2 x + 1)}^{\frac{1}{3} \cdot 3} = 20^{3}$

Rewrite the expression.

$64 {(2 x + 1)}^{1} = 20^{3}$

Simplify.

$64 (2 x + 1) = 20^{3}$

Apply the distributive property.

$64 (2 x) + 64 \cdot 1 = 20^{3}$

Multiply.

Multiply $2$ by $64$ .

$128 x + 64 \cdot 1 = 20^{3}$

Multiply $64$ by $1$ .

$128 x + 64 = 20^{3}$

Simplify the right side.

Raise $20$ to the power of $3$ .

$128 x + 64 = 8000$

Step 3

Solve for $x$ .

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

Subtract $64$ from both sides of the equation.

$128 x = 8000 - 64$

Subtract $64$ from $8000$ .

$128 x = 7936$

Divide each term in $128 x = 7936$ by $128$ and simplify.

Divide each term in $128 x = 7936$ by $128$ .

$\frac{128 x}{128} = \frac{7936}{128}$

Simplify the left side.

Cancel the common factor of $128$ .

Cancel the common factor.

$\frac{128 x}{128} = \frac{7936}{128}$

Divide $x$ by $1$ .

$x = \frac{7936}{128}$

Simplify the right side.

Divide $7936$ by $128$ .

$x = 62$

Solve for x 4 cube root of 2x+1=20

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