Pre-Algebra Examples

$\sqrt[3]{24} + \sqrt{80} + 6 \sqrt{40}$

Step 1

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

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Step 1.1

Factor $8$ out of $24$ .

$\sqrt[3]{8 (3)} + \sqrt{80} + 6 \sqrt{40}$

Step 1.2

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

$\sqrt[3]{2^{3} \cdot 3} + \sqrt{80} + 6 \sqrt{40}$

Step 2

Pull terms out from under the radical.

$2 \sqrt[3]{3} + \sqrt{80} + 6 \sqrt{40}$

Step 3

Rewrite $80$ as $4^{2} \cdot 5$ .

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Step 3.1

Factor $16$ out of $80$ .

$2 \sqrt[3]{3} + \sqrt{16 (5)} + 6 \sqrt{40}$

Step 3.2

Rewrite $16$ as $4^{2}$ .

$2 \sqrt[3]{3} + \sqrt{4^{2} \cdot 5} + 6 \sqrt{40}$

Step 4

Pull terms out from under the radical.

$2 \sqrt[3]{3} + 4 \sqrt{5} + 6 \sqrt{40}$

Step 5

Rewrite $40$ as $2^{2} \cdot 10$ .

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Step 5.1

Factor $4$ out of $40$ .

$2 \sqrt[3]{3} + 4 \sqrt{5} + 6 \sqrt{4 (10)}$

Step 5.2

Rewrite $4$ as $2^{2}$ .

$2 \sqrt[3]{3} + 4 \sqrt{5} + 6 \sqrt{2^{2} \cdot 10}$

Step 6

Pull terms out from under the radical.

$2 \sqrt[3]{3} + 4 \sqrt{5} + 6 (2 \sqrt{10})$

Step 7

Multiply $2$ by $6$ .

$2 \sqrt[3]{3} + 4 \sqrt{5} + 12 \sqrt{10}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$2 \sqrt[3]{3} + 4 \sqrt{5} + 12 \sqrt{10}$

Decimal Form:

$49.77610297 \dots$