Question

question
simplify \\(sqrt3{64x^{27}}\\) completely given \\(x > 0\\).

answer attempt 1 out of 2

Explanation:

Step1: Recall cube root properties

The cube root of a product is the product of the cube roots, i.e., $\sqrt[3]{ab}=\sqrt[3]{a}\cdot\sqrt[3]{b}$ for real numbers $a,b$. So we can split $\sqrt[3]{64x^{27}}$ into $\sqrt[3]{64}\cdot\sqrt[3]{x^{27}}$.

Step2: Simplify $\sqrt[3]{64}$

We know that $4^3 = 64$, so $\sqrt[3]{64}=\sqrt[3]{4^3}=4$.

Step3: Simplify $\sqrt[3]{x^{27}}$

Using the property of exponents for roots, $\sqrt[n]{a^m}=a^{\frac{m}{n}}$. Here $n = 3$ and $m = 27$, so $\sqrt[3]{x^{27}}=x^{\frac{27}{3}}=x^9$.

Step4: Multiply the results

Multiply the results from Step2 and Step3: $4\cdot x^9 = 4x^9$.

Answer:

$4x^9$