# How do you multiply (x^(1/3)+x^(-1/3))^2?

Sep 6, 2016

${x}^{\frac{2}{3}} + 2 + {x}^{- \frac{2}{3}}$

#### Explanation:

${\left({x}^{\frac{1}{3}} {+}^{- \frac{1}{3}}\right)}^{2} = \left({x}^{\frac{1}{3}} + {x}^{- \frac{1}{3}}\right) \left({x}^{\frac{1}{3}} + {x}^{- \frac{1}{3}}\right)$

We must ensure when multiplying that each term in the 2nd bracket is multiplied by each term in the 1st bracket.
This can be done as follows.

$\left(\textcolor{red}{{x}^{\frac{1}{3}} + {x}^{- \frac{1}{3}}}\right) \left({x}^{\frac{1}{3}} + {x}^{- \frac{1}{3}}\right)$

$= \textcolor{red}{{x}^{\frac{1}{3}}} \left({x}^{\frac{1}{3}} + {x}^{- \frac{1}{3}}\right) \textcolor{red}{+ {x}^{- \frac{1}{3}}} \left({x}^{\frac{1}{3}} + {x}^{- \frac{1}{3}}\right)$

now distribute the brackets
$\textcolor{b l u e}{\text{--------------------------------------------------------------}}$

We require to use the $\textcolor{b l u e}{\text{laws of exponents}}$

color(orange)"Reminder " color(red)(|bar(ul(color(white)(a/a)color(black)(a^mxxa^n=a^(m+n)" and " a^0=1)color(white)(a/a)|)))
$\textcolor{b l u e}{\text{------------------------------------------------------------------}}$

$= {x}^{\frac{1}{3} + \frac{1}{3}} + {x}^{\frac{1}{3} - \frac{1}{3}} + {x}^{- \frac{1}{3} + \frac{1}{3}} + {x}^{- \frac{1}{3} - \frac{1}{3}}$

$= {x}^{\frac{2}{3}} + {x}^{0} + {x}^{0} + {x}^{- \frac{2}{3}}$

$= {x}^{\frac{2}{3}} + 1 + 1 + {x}^{- \frac{2}{3}} = {x}^{\frac{2}{3}} + 2 + {x}^{- \frac{2}{3}}$