How do I reduce the index of a radical $\sqrt[9]{64a^6b^{12}}$?

Question

$\sqrt[9]{64a^6b^{12}}$

It doesn't have to be the answer, just tell me what I should do

Answer 1

$64= 2^6$ so this could be written $(2^6a^6b^{12})^{1/9}= 2^{6/9}a^{6/9}b^{12/9}= 2^{2/3}a^{2/3}b^{4/3}= b\sqrt[3]{2^2a^2b}$.

Answer 2

To reduce $(a)^{(1/b)}$ where $a,b \,\in \,\mathbb{Z^+}$ and $1 < a,b$ the first step is to try to factor $b$.

Your goal is to try to find positive integers $r,s$ such that the following is true:

• $(r \times s) = b.$

• There exists a positive integer $c$ such that $a = c^r$.

Then, you have that

$$(a)^{(1/b)} = \left(c^r\right)^{\frac{1}{(r\times s)}} = \left\{\left(c^r\right)^{1/r}\right\}^{1/s} = (c)^{1/s}.$$

Answer 3

Notice that, $\sqrt[n]{x}=x^{\frac{1}{n}}$, in your case, $n = 9$, and $64=2^6$, so

$$\sqrt[9]{64a^6b^{12}}=\sqrt[9]{2^6a^6b^{12}}=\sqrt[9]{(2a)^6b^{12}}$$

Using the above property,

$$\sqrt[9]{(2a)^6b^{12}}=\Big((2a)^6b^{12}\Big)^{\frac{1}{9}}=(2a)^{\frac{6}{9}}b^{\frac{12}{9}}=(2a)^{\frac{2}{3}}b^{\frac{4}{3}}=\Big((2a)^2b^{4}\Big)^{\frac{1}{3}}$$

Therefore,

$$\sqrt[9]{(2a)^6b^{12}}=\Big((2a)^2b^{4}\Big)^{\frac{1}{3}}=\sqrt[3]{(2a)^2b^{4}}$$

It is possible to simplify a little more, noting that $b^4=b\cdot b^3$, so

$$\sqrt[3]{(2a)^2b^{4}}=\sqrt[3]{(2a)^2b\cdot b^3}=b\sqrt[3]{(2a)^2b}$$