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

The solution to expand-((2-2x)^3)/3 is (8x^3)/3-8x^2+8x-8/3

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

expand

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

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

Solution steps

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

(2 - 2 x)^{3} = 8 - 24 x + 24 x^{2} - 8 x^{3}

= - \frac{- 8 x ^{3} + 24 x ^{2} - 24 x + 8}{3}

Apply the fraction rule:

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

\frac{8 - 24 x + 24 x ^{2} - 8 x ^{3}}{3} = - (\frac{8}{3}) - (- \frac{24 x}{3}) - (\frac{24 x ^{2}}{3}) - (- \frac{8 x ^{3}}{3})

= - (\frac{8}{3}) - (- \frac{24 x}{3}) - (\frac{24 x ^{2}}{3}) - (- \frac{8 x ^{3}}{3})

Remove parentheses:

(a) = a, - (- a) = a

= - \frac{8}{3} + \frac{24 x}{3} - \frac{24 x ^{2}}{3} + \frac{8 x ^{3}}{3}

Divide the numbers:

\frac{24}{3} = 8

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

Rewrite in standard form

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

\frac{1}{( 2 x + 7 ) ^{9}}

(2 m L - k + t^{2})^{2}

z + 0.016 k \cdot (z - 1) \cdot (1 - e^{- 0.5 \cdot T})

(\frac{cos ( t )}{t} - sin (t) ln (t))^{2}

\frac{7 x ^{3} + 9 x ^{2}}{( x ^{2} + 8 ) ^{2}}

What is expand-((2-2x)^3)/3 ?
The solution to expand-((2-2x)^3)/3 is (8x^3)/3-8x^2+8x-8/3