Solve for x 2 cube root of x-10+4=0

$2 \sqrt[3]{x - 10} + 4 = 0$

Step 1

Subtract $4$ from both sides of the equation.

$2 \sqrt[3]{x - 10} = - 4$

Step 2

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

${(2 \sqrt[3]{x - 10})}^{3} = {(- 4)}^{3}$

Step 3

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x - 10}$ as ${(x - 10)}^{\frac{1}{3}}$ .

${(2 {(x - 10)}^{\frac{1}{3}})}^{3} = {(- 4)}^{3}$

Simplify the left side.

Simplify ${(2 {(x - 10)}^{\frac{1}{3}})}^{3}$ .

Apply the product rule to $2 {(x - 10)}^{\frac{1}{3}}$ .

$2^{3} {({(x - 10)}^{\frac{1}{3}})}^{3} = {(- 4)}^{3}$

Raise $2$ to the power of $3$ .

$8 {({(x - 10)}^{\frac{1}{3}})}^{3} = {(- 4)}^{3}$

Multiply the exponents in ${({(x - 10)}^{\frac{1}{3}})}^{3}$ .

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

$8 {(x - 10)}^{\frac{1}{3} \cdot 3} = {(- 4)}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$8 {(x - 10)}^{\frac{1}{3} \cdot 3} = {(- 4)}^{3}$

Rewrite the expression.

$8 {(x - 10)}^{1} = {(- 4)}^{3}$

Simplify.

$8 (x - 10) = {(- 4)}^{3}$

Apply the distributive property.

$8 x + 8 \cdot - 10 = {(- 4)}^{3}$

Multiply $8$ by $- 10$ .

$8 x - 80 = {(- 4)}^{3}$

Simplify the right side.

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

$8 x - 80 = - 64$

Step 4

Solve for $x$ .

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

Add $80$ to both sides of the equation.

$8 x = - 64 + 80$

Add $- 64$ and $80$ .

$8 x = 16$

Divide each term in $8 x = 16$ by $8$ and simplify.

Divide each term in $8 x = 16$ by $8$ .

$\frac{8 x}{8} = \frac{16}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 x}{8} = \frac{16}{8}$

Divide $x$ by $1$ .

$x = \frac{16}{8}$

Simplify the right side.

Divide $16$ by $8$ .

$x = 2$

Solve for x 2 cube root of x-10+4=0

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