What is expand ((4y+x)^3)/8 ?

The solution to expand ((4y+x)^3)/8 is 8y^3+6y^2x+(3yx^2)/2+(x^3)/8

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expand ((4y+x)^3)/8

expand

\frac{( 4 y + x ) ^{3}}{8}

8 y^{3} + 6 y^{2} x + \frac{3 y x ^{2}}{2} + \frac{x ^{3}}{8}

Solution steps

\frac{( 4 y + x ) ^{3}}{8}

(4 y + x)^{3} = 64 y^{3} + 48 y^{2} x + 12 y x^{2} + x^{3}

= \frac{64 y ^{3} + 48 y ^{2} x + 12 y x ^{2} + x ^{3}}{8}

Apply the fraction rule:

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

\frac{64 y ^{3} + 48 y ^{2} x + 12 y x ^{2} + x ^{3}}{8} = \frac{64 y ^{3}}{8} + \frac{48 y ^{2} x}{8} + \frac{12 y x ^{2}}{8} + \frac{x ^{3}}{8}

= \frac{64 y ^{3}}{8} + \frac{48 y ^{2} x}{8} + \frac{12 y x ^{2}}{8} + \frac{x ^{3}}{8}

Divide the numbers:

\frac{64}{8} = 8

= 8 y^{3} + \frac{48 y ^{2} x}{8} + \frac{12 y x ^{2}}{8} + \frac{x ^{3}}{8}

Divide the numbers:

\frac{48}{8} = 6

= 8 y^{3} + 6 y^{2} x + \frac{12 y x ^{2}}{8} + \frac{x ^{3}}{8}

Cancel

\frac{12 y x ^{2}}{8} : \frac{3 y x ^{2}}{2}

= 8 y^{3} + 6 y^{2} x + \frac{3 y x ^{2}}{2} + \frac{x ^{3}}{8}

e^{x} (x + i y) (cos (y) + i sin (y))

(1 - y e^{x y}) \cdot (e^{- x y} + x)

\frac{5 e ^{- x}}{x ( x + 1 )}

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

(\frac{z - 1}{z T})^{2}

What is expand ((4y+x)^3)/8 ?
The solution to expand ((4y+x)^3)/8 is 8y^3+6y^2x+(3yx^2)/2+(x^3)/8