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

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

Answers (1)

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: \(\displaystyle{\left({x}+{1}\right)}^{{\frac{2}{{3}}}}\)
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