Question

# Simplify each expression (a) displaystyle{sqrt[{{4}}]{{{3}^{2}}}} (b) displaystyle{sqrt[{{6}}]{{{left({x}+{1}right)}^{4}}}}

Simplify each expression
(a) $$\displaystyle{\sqrt[{{4}}]{{{3}^{2}}}}$$
(b) $$\displaystyle{\sqrt[{{6}}]{{{\left({x}+{1}\right)}^{4}}}}$$

2021-01-16
a) Definition:
If $$a =\ \text{real number and}\ n =\ \text{positive integer such that the principal nth}\ \sqrt[]{}\ \text{of a exists, then}\ \displaystyle{a}^{{{1}\text{/}{n}}}\ \text{is defined as}\ \displaystyle{a}^{{{1}\text{/}{n}}}={\sqrt[{{n}}]{{a}}}$$
Hence, if m is a positive integer, then
$$\displaystyle{a}^{{{m}\text{/}{n}}}={\left({a}^{{{1}\text{/}{n}}}\right)}^{m}$$
$$\displaystyle{1}\text{/}{n}\ \text{and}\ \displaystyle{m}\text{/}{n}$$ are rational exponents of a.
Calculation:
From the above definition,
$$\displaystyle{\sqrt[{{4}}]{{{3}^{2}}}}$$
$$\displaystyle={\left({3}^{2}\right)}^{{\frac{1}{{4}}}}$$
$$\displaystyle={3}^{{{2}\times\frac{1}{{4}}}}$$
$$\displaystyle={3}^{{\frac{1}{{2}}}}$$
Answer: $$\displaystyle\sqrt{{3}}$$
b) Definition:
If $$a =\ \text{real number and}\ n =\ \text{positive integer such that the principal nth}\ \sqrt[]{}\ \text{of a exists, then}\ \displaystyle{a}^{{{1}\text{/}{n}}}\ \text{is defined as}\ \displaystyle{a}^{{{1}\text{/}{n}}}={\sqrt[{{n}}]{{a}}}$$
Hence, if m is a positive integer, then
$$\displaystyle{a}^{{{m}\text{/}{n}}}={\left({a}^{{{1}\text{/}{n}}}\right)}^{m}$$
$$1/n\ \text{and}\ m/n$$ are rational exponents of a.
Calcucation:
$$\displaystyle{\sqrt[{{6}}]{{{\left({x}+{1}\right)}^{4}}}}$$
$$\displaystyle={\left({\left({x}+{1}\right)}^{4}\right)}^{{\frac{1}{{6}}}}$$
$$\displaystyle={\left({x}+{1}\right)}^{{{4}\times\frac{1}{{6}}}}$$
Answer: $$\displaystyle{\left({x}+{1}\right)}^{{\frac{2}{{3}}}}$$