Solve by Factoring 3 cube root of 12+5 cube root of 12=(3+5) cube root of 24

$3 \sqrt[3]{12} + 5 \sqrt[3]{12} = (3 + 5) \sqrt[3]{24}$

Subtract $(3 + 5) \sqrt[3]{24}$ from both sides of the equation.

$3 \sqrt[3]{12} + 5 \sqrt[3]{12} - (3 + 5) \sqrt[3]{24} = 0$

Simplify $3 \sqrt[3]{12} + 5 \sqrt[3]{12} - (3 + 5) \sqrt[3]{24}$ .

Simplify each term.

Add $3$ and $5$ .

$3 \sqrt[3]{12} + 5 \sqrt[3]{12} - 1 \cdot 8 \sqrt[3]{24} = 0$

Multiply $- 1$ by $8$ .

$3 \sqrt[3]{12} + 5 \sqrt[3]{12} - 8 \sqrt[3]{24} = 0$

Rewrite $24$ as $2^{3} \cdot 3$ .

Factor $8$ out of $24$ .

$3 \sqrt[3]{12} + 5 \sqrt[3]{12} - 8 \sqrt[3]{8 (3)} = 0$

Rewrite $8$ as $2^{3}$ .

$3 \sqrt[3]{12} + 5 \sqrt[3]{12} - 8 \sqrt[3]{2^{3} \cdot 3} = 0$

Pull terms out from under the radical.

$3 \sqrt[3]{12} + 5 \sqrt[3]{12} - 8 (2 \sqrt[3]{3}) = 0$

Multiply $2$ by $- 8$ .

$3 \sqrt[3]{12} + 5 \sqrt[3]{12} - 16 \sqrt[3]{3} = 0$

Add $3 \sqrt[3]{12}$ and $5 \sqrt[3]{12}$ .

$8 \sqrt[3]{12} - 16 \sqrt[3]{3} = 0$

Since $8 \sqrt[3]{12} - 16 \sqrt[3]{3} \neq 0$ , there are no solutions.

No solution