Algebra Examples

$- 5 \sqrt[3]{14 p} \cdot - 4 \sqrt[3]{4 p^{3}}$

Step 1

Multiply $- 4$ by $- 5$ .

$20 \sqrt[3]{14 p} \sqrt[3]{4 p^{3}}$

Step 2

Reorder $4$ and $p^{3}$ .

$20 \sqrt[3]{14 p} \sqrt[3]{p^{3} \cdot 4}$

Step 3

Pull terms out from under the radical.

$20 \sqrt[3]{14 p} (p \sqrt[3]{4})$

Step 4

Multiply $20 \sqrt[3]{14 p} (p \sqrt[3]{4})$ .

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Step 4.1

Combine using the product rule for radicals.

$20 (p \sqrt[3]{14 p \cdot 4})$

Step 4.2

Multiply $4$ by $14$ .

$20 (p \sqrt[3]{56 p})$

$20 (p \sqrt[3]{56 p})$

Step 5

Rewrite $56 p$ as $2^{3} \cdot (7 p)$ .

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Step 5.1

Factor $8$ out of $56$ .

$20 (p \sqrt[3]{8 (7) p})$

Step 5.2

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

$20 (p \sqrt[3]{2^{3} \cdot 7 p})$

Step 5.3

Add parentheses.

$20 (p \sqrt[3]{2^{3} \cdot (7 p)})$

$20 (p \sqrt[3]{2^{3} \cdot (7 p)})$

Step 6

Pull terms out from under the radical.

$20 (p (2 \sqrt[3]{7 p}))$

Step 7

Rewrite using the commutative property of multiplication.

$20 (2 p \sqrt[3]{7 p})$

Step 8

Multiply $2$ by $20$ .

$40 p \sqrt[3]{7 p}$

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$