What is expand 4((3(k+1)^3)/8-((k+1)/2))+(k+1)^3 ?

The solution to expand 4((3(k+1)^3)/8-((k+1)/2))+(k+1)^3 is k^3+3k^2+(3(k+1)^3)/2+k-1

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expand 4((3(k+1)^3)/8-((k+1)/2))+(k+1)^3

expand

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

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

Solution steps

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

Simplify

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

= 4 (\frac{3 ( k + 1 ) ^{3}}{8} - \frac{k + 1}{2}) + (k + 1)^{3}

Expand

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

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

Expand

(k + 1)^{3} : k^{3} + 3 k^{2} + 3 k + 1

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

Simplify

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

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

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

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2 (sin (t)) (t) (7 - t^{2})

What is expand 4((3(k+1)^3)/8-((k+1)/2))+(k+1)^3 ?
The solution to expand 4((3(k+1)^3)/8-((k+1)/2))+(k+1)^3 is k^3+3k^2+(3(k+1)^3)/2+k-1