What is expand ((4z-1)^2(3z+1))/(48) ?

The solution to expand ((4z-1)^2(3z+1))/(48) is z^3-(z^2)/6-(5z)/(48)+1/48

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expand ((4z-1)^2(3z+1))/(48)

expand

\frac{( 4 z - 1 ) ^{2} ( 3 z + 1 )}{48}

z^{3} - \frac{z ^{2}}{6} - \frac{5 z}{48} + \frac{1}{48}

Solution steps

\frac{( 4 z - 1 ) ^{2} ( 3 z + 1 )}{48}

(4 z - 1)^{2} = 16 z^{2} - 8 z + 1

= \frac{( 16 z ^{2} - 8 z + 1 ) ( 3 z + 1 )}{48}

Expand

(16 z^{2} - 8 z + 1) (3 z + 1) : 48 z^{3} - 8 z^{2} - 5 z + 1

= \frac{48 z ^{3} - 8 z ^{2} - 5 z + 1}{48}

Apply the fraction rule:

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

\frac{48 z ^{3} - 8 z ^{2} - 5 z + 1}{48} = \frac{48 z ^{3}}{48} - \frac{8 z ^{2}}{48} - \frac{5 z}{48} + \frac{1}{48}

= \frac{48 z ^{3}}{48} - \frac{8 z ^{2}}{48} - \frac{5 z}{48} + \frac{1}{48}

Divide the numbers:

\frac{48}{48} = 1

= z^{3} - \frac{8 z ^{2}}{48} - \frac{5 z}{48} + \frac{1}{48}

Cancel

\frac{8 z ^{2}}{48} : \frac{z ^{2}}{6}

= z^{3} - \frac{z ^{2}}{6} - \frac{5 z}{48} + \frac{1}{48}

\frac{e ^{- 2 - 2} ( 3 + 2 2 )}{50}

u (2, - 3, 1) v (- 3, 1.2)

cos (4 t) u (t - π)

\frac{d}{d y} ((x) (3 x + 5 y))

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

What is expand ((4z-1)^2(3z+1))/(48) ?
The solution to expand ((4z-1)^2(3z+1))/(48) is z^3-(z^2)/6-(5z)/(48)+1/48