What is expand-(3u^3(4y^3-5))/(2y^2) ?

The solution to expand-(3u^3(4y^3-5))/(2y^2) is -6u^3y+(15u^3)/(2y^2)

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expand-(3u^3(4y^3-5))/(2y^2)

expand

- \frac{3 u ^{3} ( 4 y ^{3} - 5 )}{2 y ^{2}}

- 6 u^{3} y + \frac{15 u ^{3}}{2 y ^{2}}

Solution steps

- \frac{3 u ^{3} ( 4 y ^{3} - 5 )}{2 y ^{2}}

Expand

3 u^{3} (4 y^{3} - 5) : 12 u^{3} y^{3} - 15 u^{3}

= - \frac{12 u ^{3} y ^{3} - 15 u ^{3}}{2 y ^{2}}

Apply the fraction rule:

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

\frac{12 u ^{3} y ^{3} - 15 u ^{3}}{2 y ^{2}} = - (\frac{12 u ^{3} y ^{3}}{2 y ^{2}}) - (- \frac{15 u ^{3}}{2 y ^{2}})

= - (\frac{12 u ^{3} y ^{3}}{2 y ^{2}}) - (- \frac{15 u ^{3}}{2 y ^{2}})

Remove parentheses:

(a) = a, - (- a) = a

= - \frac{12 u ^{3} y ^{3}}{2 y ^{2}} + \frac{15 u ^{3}}{2 y ^{2}}

Cancel

\frac{12 u ^{3} y ^{3}}{2 y ^{2}} : 6 u^{3} y

= - 6 u^{3} y + \frac{15 u ^{3}}{2 y ^{2}}

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

lo g_{x^{2}} (y)

\frac{1}{( s + 1 ) ( s ^{2} + 6 s + 13 )}

cos (a t) cos ((a + b) t + c)

\frac{d ^{5}}{d} x^{5} (y = 8 x^{4} - 16 x^{3} - 24 x^{2} + 32 x + 8)

What is expand-(3u^3(4y^3-5))/(2y^2) ?
The solution to expand-(3u^3(4y^3-5))/(2y^2) is -6u^3y+(15u^3)/(2y^2)