Simplify cube root of (8y^3)^4 sixth root of (8y^3)^4

$\sqrt[3]{{(8 y^{3})}^{4}} \sqrt[6]{{(8 y^{3})}^{4}}$

Simplify the expression.

Apply the product rule to $8 y^{3}$ .

$\sqrt[3]{8^{4} {(y^{3})}^{4}} \sqrt[6]{{(8 y^{3})}^{4}}$

Raise $8$ to the power of $4$ .

$\sqrt[3]{4096 {(y^{3})}^{4}} \sqrt[6]{{(8 y^{3})}^{4}}$

Multiply the exponents in ${(y^{3})}^{4}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\sqrt[3]{4096 y^{3 \cdot 4}} \sqrt[6]{{(8 y^{3})}^{4}}$

Multiply $3$ by $4$ .

$\sqrt[3]{4096 y^{12}} \sqrt[6]{{(8 y^{3})}^{4}}$

Rewrite $4096 y^{12}$ as ${(16 y^{4})}^{3}$ .

$\sqrt[3]{{(16 y^{4})}^{3}} \sqrt[6]{{(8 y^{3})}^{4}}$

Pull terms out from under the radical, assuming real numbers.

$16 y^{4} \sqrt[6]{{(8 y^{3})}^{4}}$

Simplify the expression.

Apply the product rule to $8 y^{3}$ .

$16 y^{4} \sqrt[6]{8^{4} {(y^{3})}^{4}}$

Raise $8$ to the power of $4$ .

$16 y^{4} \sqrt[6]{4096 {(y^{3})}^{4}}$

Multiply the exponents in ${(y^{3})}^{4}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$16 y^{4} \sqrt[6]{4096 y^{3 \cdot 4}}$

Multiply $3$ by $4$ .

$16 y^{4} \sqrt[6]{4096 y^{12}}$

Rewrite $4096 y^{12}$ as ${(4 y^{2})}^{6}$ .

$16 y^{4} \sqrt[6]{{(4 y^{2})}^{6}}$

Pull terms out from under the radical, assuming positive real numbers.

$16 y^{4} (4 y^{2})$

Rewrite using the commutative property of multiplication.

$16 \cdot 4 y^{4} y^{2}$

Multiply $y^{4}$ by $y^{2}$ by adding the exponents.

Move $y^{2}$ .

$16 \cdot 4 (y^{2} y^{4})$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$16 \cdot 4 y^{2 + 4}$

Add $2$ and $4$ .

$16 \cdot 4 y^{6}$

Multiply $16$ by $4$ .

$64 y^{6}$