What is expand |(((x+1)^3)/(2^x))-2| ?

The solution to expand |(((x+1)^3)/(2^x))-2| is |(x^3)/(2^x)+(3x^2)/(2^x)+(3x)/(2^x)+1/(2^x)-2|

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expand |(((x+1)^3)/(2^x))-2|

expand

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

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

Solution steps

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

Remove parentheses:

(a) = a

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

(x + 1)^{3} = x^{3} + 3 x^{2} + 3 x + 1

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

Apply the fraction rule:

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

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

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

(x + 2 \cdot lo g_{2} (x)) (x - 2)

(x - 1) e^{3 x + 2}

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

(2 u, 2 v, 1) \cdot (9 - u^{2} - v^{2}, 9 - u^{2} - v^{2}, u)

\frac{n ^{2}}{( - 2 i d x + 2 i ^{2} d ^{2} ) ( s + x - i d )}

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