Question

solve:
$2\sqrt3{3x - 1} - 6 = -2$
$x = \square$

Explanation:

Step1: Isolate the cube root term

Add 6 to both sides.
$2\sqrt[3]{3x - 1} = -2 + 6$
$2\sqrt[3]{3x - 1} = 4$

Step2: Simplify the equation

Divide both sides by 2.
$\sqrt[3]{3x - 1} = \frac{4}{2}$
$\sqrt[3]{3x - 1} = 2$

Step3: Eliminate the cube root

Cube both sides of the equation.
$(\sqrt[3]{3x - 1})^3 = 2^3$
$3x - 1 = 8$

Step4: Solve for x

Add 1 to both sides, then divide by 3.
$3x = 8 + 1$
$3x = 9$
$x = \frac{9}{3}$

Answer:

$x = 3$