What is expand ((a^{2n}-1)^2)/(a^{2n)} ?

The solution to expand ((a^{2n}-1)^2)/(a^{2n)} is a^{2n}-2+1/(a^{2n)}

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expand ((a^{2n}-1)^2)/(a^{2n)}

expand

\frac{( a ^{2 n} - 1 ) ^{2}}{a ^{2 n}}

a^{2 n} - 2 + \frac{1}{a ^{2 n}}

Solution steps

\frac{( a ^{2 n} - 1 ) ^{2}}{a ^{2 n}}

(a^{2 n} - 1)^{2} = a^{4 n} - 2 a^{2 n} + 1

= \frac{a ^{4 n} - 2 a ^{2 n} + 1}{a ^{2 n}}

Apply the fraction rule:

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

\frac{a ^{4 n} - 2 a ^{2 n} + 1}{a ^{2 n}} = \frac{a ^{4 n}}{a ^{2 n}} - \frac{2 a ^{2 n}}{a ^{2 n}} + \frac{1}{a ^{2 n}}

= \frac{a ^{4 n}}{a ^{2 n}} - \frac{2 a ^{2 n}}{a ^{2 n}} + \frac{1}{a ^{2 n}}

Cancel

\frac{a ^{4 n}}{a ^{2 n}} : a^{2 n}

= a^{2 n} - \frac{2 a ^{2 n}}{a ^{2 n}} + \frac{1}{a ^{2 n}}

Cancel

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

= a^{2 n} - 2 + \frac{1}{a ^{2 n}}

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

\frac{( k ( k + 1 ) ( 2 k + 1 ) )}{6}

\frac{1}{( x ^{2} - b ^{2} ) ^{2}}

(9 - 2 x)^{2} (x) 0

(A - B) \circ - B

What is expand ((a^{2n}-1)^2)/(a^{2n)} ?
The solution to expand ((a^{2n}-1)^2)/(a^{2n)} is a^{2n}-2+1/(a^{2n)}