Simplify fourth root of 256x^2* cube root of 27x

$\sqrt[4]{256 x^{2}} \cdot \sqrt[3]{27 x}$

Rewrite $256$ as $4^{4}$ .

$\sqrt[4]{4^{4} x^{2}} \cdot \sqrt[3]{27 x}$

Pull terms out from under the radical.

$4 \sqrt[4]{x^{2}} \cdot \sqrt[3]{27 x}$

Rewrite $\sqrt[4]{x^{2}}$ as $\sqrt{\sqrt{x^{2}}}$ .

$4 \sqrt{\sqrt{x^{2}}} \cdot \sqrt[3]{27 x}$

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

$4 \sqrt{x} \cdot \sqrt[3]{27 x}$

Rewrite $27$ as $3^{3}$ .

$4 \sqrt{x} \cdot \sqrt[3]{3^{3} x}$

Pull terms out from under the radical.

$4 \sqrt{x} \cdot (3 \sqrt[3]{x})$

Multiply $4 \sqrt{x} (3 \sqrt[3]{x})$ .

Multiply $3$ by $4$ .

$12 \sqrt{x} \sqrt[3]{x}$

Rewrite the expression using the least common index of $6$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x}$ as $x^{\frac{1}{3}}$ .

$12 (x^{\frac{1}{3}} \sqrt{x})$

Rewrite $x^{\frac{1}{3}}$ as $x^{\frac{2}{6}}$ .

$12 (x^{\frac{2}{6}} \sqrt{x})$

Rewrite $x^{\frac{2}{6}}$ as $\sqrt[6]{x^{2}}$ .

$12 (\sqrt[6]{x^{2}} \sqrt{x})$

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

$12 (\sqrt[6]{x^{2}} x^{\frac{1}{2}})$

Rewrite $x^{\frac{1}{2}}$ as $x^{\frac{3}{6}}$ .

$12 (\sqrt[6]{x^{2}} x^{\frac{3}{6}})$

Rewrite $x^{\frac{3}{6}}$ as $\sqrt[6]{x^{3}}$ .

$12 (\sqrt[6]{x^{2}} \sqrt[6]{x^{3}})$

Combine using the product rule for radicals.

$12 \sqrt[6]{x^{2} x^{3}}$

Multiply $x^{2}$ by $x^{3}$ by adding the exponents.

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

$12 \sqrt[6]{x^{2 + 3}}$

Add $2$ and $3$ .

$12 \sqrt[6]{x^{5}}$