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

The solution to expand x^2(2-x)+1/3 (2-x)^3+(2-x) is 4x^2-(4x^3)/3+14/3-5x

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

expand

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

4 x^{2} - \frac{4 x ^{3}}{3} + \frac{14}{3} - 5 x

Solution steps

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

Apply rule:

a + (b + c) = a + b + c

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

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

Expand

x^{2} (2 - x) : 2 x^{2} - x^{3}

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

Expand

\frac{1}{3} (2 - x)^{3} : \frac{8}{3} - 4 x + 2 x^{2} - \frac{x ^{3}}{3}

= 2 x^{2} - x^{3} + \frac{8}{3} - 4 x + 2 x^{2} - \frac{x ^{3}}{3} + 2 - x

Simplify

2 x^{2} - x^{3} + \frac{8}{3} - 4 x + 2 x^{2} - \frac{x ^{3}}{3} + 2 - x : 4 x^{2} - \frac{4 x ^{3}}{3} + \frac{14}{3} - 5 x

= 4 x^{2} - \frac{4 x ^{3}}{3} + \frac{14}{3} - 5 x

t (- 9 cos (u) (a t) - 4 sin (a t))

\frac{1}{4 \cdot π \cdot 0.02} (\frac{1}{0.75} - \frac{1}{0.85}) + \frac{1}{5}

\frac{( 6 x ^{2} - 2 )}{( x ^{2} + 1 ) ^{3}}

x - 3 (x + 3)

\frac{1}{( x + 4 ) ( x ^{3} + 6 x ^{2} + 6 )}

What is expand x^2(2-x)+1/3 (2-x)^3+(2-x) ?
The solution to expand x^2(2-x)+1/3 (2-x)^3+(2-x) is 4x^2-(4x^3)/3+14/3-5x