What is expand ((e^{2x}-1)^2)/(e^x) ?

The solution to expand ((e^{2x}-1)^2)/(e^x) is e^{3x}-2e^x+1/(e^x)

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expand ((e^{2x}-1)^2)/(e^x)

expand

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

e^{3 x} - 2 e^{x} + \frac{1}{e ^{x}}

Solution steps

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

(e^{2 x} - 1)^{2} = e^{4 x} - 2 e^{2 x} + 1

= \frac{e ^{4 x} - 2 e ^{2 x} + 1}{e ^{x}}

Apply the fraction rule:

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

\frac{e ^{4 x} - 2 e ^{2 x} + 1}{e ^{x}} = \frac{e ^{4 x}}{e ^{x}} - \frac{2 e ^{2 x}}{e ^{x}} + \frac{1}{e ^{x}}

= \frac{e ^{4 x}}{e ^{x}} - \frac{2 e ^{2 x}}{e ^{x}} + \frac{1}{e ^{x}}

Cancel

\frac{e ^{4 x}}{e ^{x}} : e^{3 x}

= e^{3 x} - \frac{2 e ^{2 x}}{e ^{x}} + \frac{1}{e ^{x}}

Cancel

\frac{2 e ^{2 x}}{e ^{x}} : 2 e^{x}

= e^{3 x} - 2 e^{x} + \frac{1}{e ^{x}}

\frac{9}{3 - 6}

π (0.5 x + 0.4 0.5 x + 5 + 5.04)

\frac{1}{( x - 1 ) ( x + 6 )}

\frac{4 - x}{( x - 3 ) ^{2}}

\frac{3 ( 1 + x )}{3 + x}

What is expand ((e^{2x}-1)^2)/(e^x) ?
The solution to expand ((e^{2x}-1)^2)/(e^x) is e^{3x}-2e^x+1/(e^x)