Solve the Inequality for x cube root of x+7>=3

$\sqrt[3]{x + 7} \geq 3$

Step 1

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

${\sqrt[3]{x + 7}}^{3} \geq 3^{3}$

Step 2

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x + 7}$ as ${(x + 7)}^{\frac{1}{3}}$ .

${({(x + 7)}^{\frac{1}{3}})}^{3} \geq 3^{3}$

Simplify the left side.

Simplify ${({(x + 7)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(x + 7)}^{\frac{1}{3}})}^{3}$ .

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

${(x + 7)}^{\frac{1}{3} \cdot 3} \geq 3^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(x + 7)}^{\frac{1}{3} \cdot 3} \geq 3^{3}$

Rewrite the expression.

${(x + 7)}^{1} \geq 3^{3}$

Simplify.

$x + 7 \geq 3^{3}$

Simplify the right side.

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

$x + 7 \geq 27$

Step 3

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

Subtract $7$ from both sides of the inequality.

$x \geq 27 - 7$

Subtract $7$ from $27$ .

$x \geq 20$

Step 4

The result can be shown in multiple forms.

Inequality Form:

$x \geq 20$

Interval Notation:

$[20, \infty)$

Step 5