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

The solution to expand (4(k+2)^2(k+1))/3 is (4k^3)/3+(20k^2)/3+(32k)/3+16/3

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

expand

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

\frac{4 k ^{3}}{3} + \frac{20 k ^{2}}{3} + \frac{32 k}{3} + \frac{16}{3}

Solution steps

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

(k + 2)^{2} = k^{2} + 4 k + 4

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

Expand

4 (k^{2} + 4 k + 4) (k + 1) : 4 k^{3} + 20 k^{2} + 32 k + 16

= \frac{4 k ^{3} + 20 k ^{2} + 32 k + 16}{3}

Apply the fraction rule:

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

\frac{4 k ^{3} + 20 k ^{2} + 32 k + 16}{3} = \frac{4 k ^{3}}{3} + \frac{20 k ^{2}}{3} + \frac{32 k}{3} + \frac{16}{3}

= \frac{4 k ^{3}}{3} + \frac{20 k ^{2}}{3} + \frac{32 k}{3} + \frac{16}{3}

\frac{d}{d t} ((x) (6 t^{2} - t^{3}))

\frac{d}{d x} ((y^{3}) c (x - c)^{2})

(x^{2} - 2 x - 3)^{5}

(8 x - 20) \circ

\frac{4 x}{( 79 - x ) ( 21 - x )}

What is expand (4(k+2)^2(k+1))/3 ?
The solution to expand (4(k+2)^2(k+1))/3 is (4k^3)/3+(20k^2)/3+(32k)/3+16/3