Evaluate 6 cube root of 3+3 cube root of 24

$6 \sqrt[3]{3} + 3 \sqrt[3]{24}$

Step 1

Simplify each term.

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

Factor $8$ out of $24$ .

$6 \sqrt[3]{3} + 3 \sqrt[3]{8 (3)}$

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

$6 \sqrt[3]{3} + 3 \sqrt[3]{2^{3} \cdot 3}$

$6 \sqrt[3]{3} + 3 \sqrt[3]{2^{3} \cdot 3}$

Pull terms out from under the radical.

$6 \sqrt[3]{3} + 3 (2 \sqrt[3]{3})$

Multiply $2$ by $3$ .

$6 \sqrt[3]{3} + 6 \sqrt[3]{3}$

$6 \sqrt[3]{3} + 6 \sqrt[3]{3}$

Step 2

Add $6 \sqrt[3]{3}$ and $6 \sqrt[3]{3}$ .

$12 \sqrt[3]{3}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$12 \sqrt[3]{3}$

Decimal Form:

$17.30699484 \dots$

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