What is expand (x^3y(9x^2+72xy-4y^2))/9 ?

The solution to expand (x^3y(9x^2+72xy-4y^2))/9 is x^5y+8x^4y^2-(4x^3y^3)/9

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expand (x^3y(9x^2+72xy-4y^2))/9

expand

\frac{x ^{3} y ( 9 x ^{2} + 72 x y - 4 y ^{2} )}{9}

x^{5} y + 8 x^{4} y^{2} - \frac{4 x ^{3} y ^{3}}{9}

Solution steps

\frac{x ^{3} y ( 9 x ^{2} + 72 x y - 4 y ^{2} )}{9}

Expand

x^{3} y (9 x^{2} + 72 x y - 4 y^{2}) : 9 x^{5} y + 72 x^{4} y^{2} - 4 x^{3} y^{3}

= \frac{9 x ^{5} y + 72 x ^{4} y ^{2} - 4 x ^{3} y ^{3}}{9}

Apply the fraction rule:

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

\frac{9 x ^{5} y + 72 x ^{4} y ^{2} - 4 x ^{3} y ^{3}}{9} = \frac{9 x ^{5} y}{9} + \frac{72 x ^{4} y ^{2}}{9} - \frac{4 x ^{3} y ^{3}}{9}

= \frac{9 x ^{5} y}{9} + \frac{72 x ^{4} y ^{2}}{9} - \frac{4 x ^{3} y ^{3}}{9}

Divide the numbers:

\frac{9}{9} = 1

= x^{5} y + \frac{72 x ^{4} y ^{2}}{9} - \frac{4 x ^{3} y ^{3}}{9}

Divide the numbers:

\frac{72}{9} = 8

= x^{5} y + 8 x^{4} y^{2} - \frac{4 x ^{3} y ^{3}}{9}

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

\frac{1125}{s ( s + 5 ) + 1125}

1.66 z (z - e^{- T}) (z - e^{- 0.6 T})

(- 2, - 3) y (4, 2)

(x^{2} + (y - c)^{2}) d x

What is expand (x^3y(9x^2+72xy-4y^2))/9 ?
The solution to expand (x^3y(9x^2+72xy-4y^2))/9 is x^5y+8x^4y^2-(4x^3y^3)/9