# 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) $$\displaystyle{\sqrt[{{6}}]{{{\left({x}+{1}\right)}^{4}}}}$$

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

### Relevant Questions

$$\displaystyle{\left({a}\right)}{\sqrt[{{6}}]{{{x}^{5}}}}$$
To simplify:
The expression $$\displaystyle{\sqrt[{{6}}]{{{x}^{5}}}}$$ and express the answer using rational exponents.
(b) $$\displaystyle{\left(\sqrt{{x}}\right)}^{9}$$
To simplify:
The expression $$\displaystyle{\left(\sqrt{{x}}\right)}^{9}$$ and express the answer using rational exponents.
To simplify:
The given redical:
An expression: $$\displaystyle{\sqrt[{{4}}]{{{\left({x}^{2}-{4}\right)}^{4}}}}.$$
Use absolute value bars when necessary.
Rational exponents evaluate each expression.
(a) $$\displaystyle{27}^{{{1}\text{/}{3}}}$$
(b) $$\displaystyle{\left(-{8}\right)}^{{{1}\text{/}{3}}}$$
(c) $$\displaystyle-{\left(\frac{1}{{8}}\right)}^{{{1}\text{/}{3}}}$$
a) Find the rational zeros and then the other zeros of the polynomial function $$\displaystyle{\left({x}\right)}={x}^{3}-{4}{x}^{2}+{2}{x}+{4},\ \tet{that is, solve}\ \displaystyle f{{\left({x}\right)}}={0}.$$
b) Factor $$f(x)$$ into linear factors.
To multiply:
The given expression. Then simplify if possible. Assume that all variables represent positive real numbers.
Given:
An expression: $$\displaystyle\sqrt{{3}}{\left(\sqrt{{27}}-\sqrt{{3}}\right)}$$
Radical simplify the expression, and eliminate any negative exponents(s). Assume that all letters denote positive numbers.
a) $$\displaystyle\sqrt{{{6}}}{\left\lbrace{y}^{{{5}}}\right\rbrace}\sqrt{{{3}}}{\left\lbrace{y}^{{{2}}}\right\rbrace}$$
b) $$\displaystyle{\left({5}\sqrt{{{3}}}{\left\lbrace{x}\right\rbrace}\right)}{\left({2}\sqrt{{{4}}}{\left\lbrace{x}\right\rbrace}\right)}$$
$$\displaystyle \tan{{\left({{\cos}^{ -{{1}}}{5}}{x}\right)}}=?$$
$$\displaystyle{\sqrt[{{4}}]{{{c}{d}^{2}}}}\times{\sqrt[{{3}}]{{{c}^{2}{d}}}}.$$
The values of x that satisfy the equation with rational exponents $$\displaystyle{\left({x}+{5}\right)}^{{\frac{3}{{2}}}}={8}$$ and check all the proposed solutions.
$$\displaystyle{\sqrt[{{7}}]{{11}}}\times{\sqrt[{{6}}]{{13}}}$$