Question

what is the solution to the equation \\(\sqrt3{x + 4} + \sqrt3{2x + 8} = 0\\)?\
\\(\bigcirc\\) \\(x = -12\\)\
\\(\bigcirc\\) \\(x = -4\\)\
\\(\bigcirc\\) \\(x = 4\\)\
\\(\bigcirc\\) \\(x = 12\\)

Explanation:

Step1: Isolate one cube root

$\sqrt[3]{x+4} = -\sqrt[3]{2x+8}$

Step2: Cube both sides

$(\sqrt[3]{x+4})^3 = (-\sqrt[3]{2x+8})^3$
$x+4 = -(2x+8)$

Step3: Simplify and solve for x

$x+4 = -2x-8$
$x+2x = -8-4$
$3x = -12$
$x = \frac{-12}{3} = -4$

Step4: Verify the solution

Substitute $x=-4$ into original equation:
$\sqrt[3]{-4+4}+\sqrt[3]{2(-4)+8} = \sqrt[3]{0}+\sqrt[3]{-8+8} = 0+0=0$, which matches the right-hand side.

Answer:

B. $x = -4$