Find the value of (7^-1/3)^3 ?

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Find the value of (7^-1/3)^3 ?

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Answer 1 · 2018-04-15T13:38:54

$\frac{1}{7}$ $1/7$

Explanation:

Start by getting rid of the negative exponent
To do that you have to flip the problem:
${(\frac{1}{7^{\frac{1}{3}}})}^{3}$ $(1/7^(1/3))^3$
Next multiply the exponent to the parenthesis.
If you cube $1$ $1$ it just stays $1$ $1$ and if you cube the $\frac{1}{3}$ $1/3$ it cancels because it becomes $\frac{3}{3}$ $3/3$ :
$\frac{1}{7^{\frac{3}{3}}}$ $1/7^(3/3)$
Once it becomes $\frac{3}{3}$ $3/3$ it simplifies to $1$ $1$ and $7^{1} = 7$ $7^1=7$
$\frac{1}{7}$ $1/7$

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Find the value of (7^-1/3)^3 ?

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