How do you simplify ${(x^{\frac{1}{3}} + x^{- \frac{1}{3}})}^{2}$ $(x^(1/3) + x^(-1/3))^2$ ?

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Answer 1 · 2016-06-27T22:27:03

See explanation.

A way we could simplify this is by multiplying out the expression.

${(x^{\frac{1}{3}} + x^{- (\frac{1}{3})})}^{2}$ $(x^(1/3) + x^-(1/3))^2$

Is the same as writing:

$(x^{\frac{1}{3}} + x^{- \frac{1}{3}})$ $(x^(1/3) + x^(-1/3))$ $(x^{\frac{1}{3}} + x^{- \frac{1}{3}})$ $(x^(1/3) + x^(-1/3))$

Which can multiply out to be:

$x^{\frac{2}{3}} + x^{0} + x^{0} + x^{- \frac{2}{3}}$ $x^(2/3) + x^(0) + x^(0) + x^(-2/3)$

Here we can combine like terms:

$x^{\frac{2}{3}} + 2 (x^{0}) + x^{- \frac{2}{3}}$ $x^(2/3) + 2(x^(0))+x^(-2/3)$

Which is ultimately:

$x^{\frac{2}{3}} + 2 + x^{- \frac{2}{3}}$ $x^(2/3) + 2 + x^(-2/3)$

Please note that there are many ways of getting to this simplified form.

How do you simplify (x^(1/3) + x^(-1/3))^2(x13+x−13)2 (x^(1/3) + x^(-1/3))^2?