Question

How do you simplify \[{{\left( {{x}^{\dfrac{1}{3}}}+{{x}^{-\dfrac{1}{3}}} \right)}^{2}}\]?

Answer 1

Hint: We first explain the process of exponents and indices. We find the general form of a square of sum of two numbers. Then we explain the different binary operations on exponents. We use the identities \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\] to find the simplified form of \[{{\left( {{x}^{\dfrac{1}{3}}}+{{x}^{-\dfrac{1}{3}}} \right)}^{2}}\].

Complete step by step solution:
We have been given an expression where we need to find the solution of a square of sum of two terms.
We use the identity of \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\].
We put the values for \[a={{x}^{\dfrac{1}{3}}},b={{x}^{-\dfrac{1}{3}}}\].
Therefore, \[{{\left( {{x}^{\dfrac{1}{3}}}+{{x}^{-\dfrac{1}{3}}} \right)}^{2}}={{\left( {{x}^{\dfrac{1}{3}}} \right)}^{2}}+{{\left( {{x}^{-\dfrac{1}{3}}} \right)}^{2}}+2\left( {{x}^{\dfrac{1}{3}}} \right)\left( {{x}^{-\dfrac{1}{3}}} \right)\].
Now we need to use the formulas for indices to simplify the solution.
Let the numbers be ${{a}^{m}}$ and ${{a}^{n}}$. We take multiplication of these numbers.
The indices get added. So, ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$.
The division works in an almost similar way. The indices get subtracted. So, $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$.
We also have the identity of ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$.
Therefore, \[{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{2}}={{x}^{\dfrac{2}{3}}},{{\left( {{x}^{-\dfrac{1}{3}}} \right)}^{2}}={{x}^{-\dfrac{2}{3}}},\left( {{x}^{\dfrac{1}{3}}} \right)\left( {{x}^{-\dfrac{1}{3}}} \right)={{x}^{\dfrac{1}{3}-\dfrac{1}{3}}}={{x}^{0}}=1\].
The simplified form of the expression will be
\[\begin{align}
  & {{\left( {{x}^{\dfrac{1}{3}}}+{{x}^{-\dfrac{1}{3}}} \right)}^{2}} \\
 & ={{\left( {{x}^{\dfrac{1}{3}}} \right)}^{2}}+{{\left( {{x}^{-\dfrac{1}{3}}} \right)}^{2}}+2\left( {{x}^{\dfrac{1}{3}}} \right)\left( {{x}^{-\dfrac{1}{3}}} \right) \\
 & ={{x}^{\dfrac{2}{3}}}+{{x}^{-\dfrac{2}{3}}}+2 \\
\end{align}\]
Therefore, the simplified form of \[{{\left( {{x}^{\dfrac{1}{3}}}+{{x}^{-\dfrac{1}{3}}} \right)}^{2}}\] is \[{{x}^{\dfrac{2}{3}}}+{{x}^{-\dfrac{2}{3}}}+2\].

Note:
The addition and subtraction for exponents works for taking common terms out depending on the values of the indices.
For numbers ${{a}^{m}}$ and ${{a}^{n}}$, we have ${{a}^{m}}\pm {{a}^{n}}={{a}^{m}}\left( 1\pm {{a}^{n-m}} \right)$.the relation is independent of the values of $m$ and $n$.