What is expand ((2x-2-ax)(1+a)(1-a))/(2a) ?

The solution to expand ((2x-2-ax)(1+a)(1-a))/(2a) is x/a-ax-1/a+a-x/2+(a^2x)/2

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expand ((2x-2-ax)(1+a)(1-a))/(2a)

expand

\frac{( 2 x - 2 - a x ) ( 1 + a ) ( 1 - a )}{2 a}

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

Solution steps

\frac{( 2 x - 2 - a x ) ( 1 + a ) ( 1 - a )}{2 a}

Expand

(2 x - 2 - a x) (1 + a) (1 - a) : 2 x - 2 a^{2} x - 2 + 2 a^{2} - a x + a^{3} x

= \frac{2 x - 2 a ^{2} x - 2 + 2 a ^{2} - a x + a ^{3} x}{2 a}

Apply the fraction rule:

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

\frac{2 x - 2 a ^{2} x - 2 + 2 a ^{2} - a x + a ^{3} x}{2 a} = \frac{2 x}{2 a} - \frac{2 a ^{2} x}{2 a} - \frac{2}{2 a} + \frac{2 a ^{2}}{2 a} - \frac{a x}{2 a} + \frac{a ^{3} x}{2 a}

= \frac{2 x}{2 a} - \frac{2 a ^{2} x}{2 a} - \frac{2}{2 a} + \frac{2 a ^{2}}{2 a} - \frac{a x}{2 a} + \frac{a ^{3} x}{2 a}

\frac{2 x}{2 a} = \frac{x}{a}

\frac{2 a ^{2} x}{2 a} = a x

\frac{2}{2 a} = \frac{1}{a}

\frac{2 a ^{2}}{2 a} = a

\frac{a x}{2 a} = \frac{x}{2}

\frac{a ^{3} x}{2 a} = \frac{a ^{2} x}{2}

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

t (x) x^{^{''}} + (4 t (x) - 2) x^{^{'}} + 4 x

\frac{d}{d y} ((x) (2 x + 1))

\frac{sec ^{2} ( x ) ( x ^{2} + 1 ) ^{6}}{( e ^{x} - cos ^{2} ( x ) ) ^{4}}

1 + t^{2} \cdot (1 + t^{2})

(2 x - y) (2 x + y)

What is expand ((2x-2-ax)(1+a)(1-a))/(2a) ?
The solution to expand ((2x-2-ax)(1+a)(1-a))/(2a) is x/a-ax-1/a+a-x/2+(a^2x)/2