What is expand ((x^3+4)(y^2+1))/(xy) ?

The solution to expand ((x^3+4)(y^2+1))/(xy) is x^2y+(x^2)/y+(4y)/x+4/(xy)

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expand ((x^3+4)(y^2+1))/(xy)

expand

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

x^{2} y + \frac{x ^{2}}{y} + \frac{4 y}{x} + \frac{4}{x y}

Solution steps

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

Expand

(x^{3} + 4) (y^{2} + 1) : x^{3} y^{2} + x^{3} + 4 y^{2} + 4

= \frac{x ^{3} y ^{2} + x ^{3} + 4 y ^{2} + 4}{x y}

Apply the fraction rule:

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

\frac{x ^{3} y ^{2} + x ^{3} + 4 y ^{2} + 4}{x y} = \frac{x ^{3} y ^{2}}{x y} + \frac{x ^{3}}{x y} + \frac{4 y ^{2}}{x y} + \frac{4}{x y}

= \frac{x ^{3} y ^{2}}{x y} + \frac{x ^{3}}{x y} + \frac{4 y ^{2}}{x y} + \frac{4}{x y}

Cancel

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

= x^{2} y + \frac{x ^{3}}{x y} + \frac{4 y ^{2}}{x y} + \frac{4}{x y}

Cancel

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

= x^{2} y + \frac{x ^{2}}{y} + \frac{4 y ^{2}}{x y} + \frac{4}{x y}

Cancel

\frac{4 y ^{2}}{x y} : \frac{4 y}{x}

= x^{2} y + \frac{x ^{2}}{y} + \frac{4 y}{x} + \frac{4}{x y}

(x = 0,) (y = - 9)

\frac{t ( 3 t - 2 ) ^{4}}{e ^{t}}

(4 e - d)^{2}

∣ 16 + 4 (3 q + p) ∣

π (60 ln (2) - \frac{27}{4})

What is expand ((x^3+4)(y^2+1))/(xy) ?
The solution to expand ((x^3+4)(y^2+1))/(xy) is x^2y+(x^2)/y+(4y)/x+4/(xy)