Solve by Factoring cube root of 6x-9-3=0

$\sqrt[3]{6 x - 9} - 3 = 0$

Step 1

Factor $3$ out of $6 x - 9$ .

Factor $3$ out of $6 x$ .

$\sqrt[3]{3 (2 x) - 9} - 3 = 0$

Factor $3$ out of $- 9$ .

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

Factor $3$ out of $3 (2 x) + 3 (- 3)$ .

$\sqrt[3]{3 (2 x - 3)} - 3 = 0$

Step 2

Add $3$ to both sides of the equation.

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

Step 3

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

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

Step 4

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Apply the distributive property.

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

Multiply.

Multiply $2$ by $3$ .

${(6 x + 3 \cdot - 3)}^{1} = 3^{3}$

Multiply $3$ by $- 3$ .

${(6 x - 9)}^{1} = 3^{3}$

Simplify.

$6 x - 9 = 3^{3}$

Simplify the right side.

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

$6 x - 9 = 27$

Step 5

Solve for $x$ .

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

Add $9$ to both sides of the equation.

$6 x = 27 + 9$

Add $27$ and $9$ .

$6 x = 36$

Divide each term in $6 x = 36$ by $6$ and simplify.

Divide each term in $6 x = 36$ by $6$ .

$\frac{6 x}{6} = \frac{36}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x}{6} = \frac{36}{6}$

Divide $x$ by $1$ .

$x = \frac{36}{6}$

Simplify the right side.

Divide $36$ by $6$ .

$x = 6$

Solve by Factoring cube root of 6x-9-3=0

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