Question

⁶√(4096x¹⁵y²⁶)

Explanation:

Step1: Simplify the radical coefficient

First, find the 6th root of 4096. Since $4^6 = 4096$, we have $\sqrt[6]{4096}=4$.

Step2: Simplify the $x$-term

Use the radical rule $\sqrt[n]{x^a}=x^{\frac{a}{n}}$. For $x^{15}$, $\frac{15}{6}=2+\frac{3}{6}=2+\frac{1}{2}$, so $\sqrt[6]{x^{15}}=x^2\sqrt[6]{x^3}=x^2\sqrt{x}$.

Step3: Simplify the $y$-term

For $y^{26}$, $\frac{26}{6}=4+\frac{2}{6}=4+\frac{1}{3}$, so $\sqrt[6]{y^{26}}=y^4\sqrt[6]{y^2}=y^4\sqrt[3]{y}$.

Step4: Combine all simplified terms

Multiply the results from Steps 1-3 together.

Answer:

$4x^2y^4\sqrt{x}\sqrt[3]{y}$ or written as $4x^2y^4x^{\frac{1}{2}}y^{\frac{1}{3}}=4x^{\frac{5}{2}}y^{\frac{13}{3}}$