What is (8n^4-4n^{-2})/(2n^{-2)} ?

The solution to (8n^4-4n^{-2})/(2n^{-2)} is 2(2n^6-1)

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simplify (8n^4-4n^{-2})/(2n^{-2)}

simplify

\frac{8 n ^{4} - 4 n ^{- 2}}{2 n ^{- 2}}

2 (2 n^{6} - 1)

Solution steps

\frac{8 n ^{4} - 4 n ^{- 2}}{2 n ^{- 2}}

Multiply

2 n^{- 2} : \frac{2}{n ^{2}}

= \frac{8 n ^{4} - 4 n ^{- 2}}{\frac{2}{n ^{2}}}

4 n^{- 2} = \frac{4}{n ^{2}}

= \frac{8 n ^{4} - \frac{4}{n ^{2}}}{\frac{2}{n ^{2}}}

Apply the fraction rule:

\frac{a}{\frac{b}{c}} = \frac{a \cdot c}{b}

= \frac{( 8 n ^{4} - \frac{4}{n ^{2}} ) n ^{2}}{2}

Join

8 n^{4} - \frac{4}{n ^{2}} : \frac{8 n ^{6} - 4}{n ^{2}}

= \frac{n ^{2} \frac{8 n ^{6} - 4}{n ^{2}}}{2}

Multiply

\frac{8 n ^{6} - 4}{n ^{2}} n^{2} : 8 n^{6} - 4

= \frac{8 n ^{6} - 4}{2}

Factor

8 n^{6} - 4 : 4 (2 n^{6} - 1)

= \frac{4 ( 2 n ^{6} - 1 )}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2 (2 n^{6} - 1)

8^{2} \cdot 8^{3}

\frac{4}{3} \div \frac{1}{6}

- 3 (4 w + 5) - 16 - 2 w

x - \frac{1}{2} x

12 \cdot 43

What is (8n^4-4n^{-2})/(2n^{-2)} ?
The solution to (8n^4-4n^{-2})/(2n^{-2)} is 2(2n^6-1)