To what power $(- 2)$ should be raised to get $16$ ?

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Solution

$16 = (- 2) \times (- 2) \times (- 2) \times (- 2) = (- 2)^{4}$
Power = $4$

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What should be added to $\frac{- 2}{4}$ to get $\frac{5}{16}$ ?

\frac{1}{x - 2} + \frac{1}{x + 2}

\frac{4 x^{3}}{x^{4} - 16}

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