What is expand (3(PB+9Mx+6My))/(2B^3) ?

The solution to expand (3(PB+9Mx+6My))/(2B^3) is (3P)/(2B^2)+(27Mx)/(2B^3)+(9My)/(B^3)

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expand (3(PB+9Mx+6My))/(2B^3)

expand

\frac{3 ( PB + 9 M x + 6 M y )}{2 B ^{3}}

\frac{3 P}{2 B ^{2}} + \frac{27 M x}{2 B ^{3}} + \frac{9 M y}{B ^{3}}

Solution steps

\frac{3 ( PB + 9 M x + 6 M y )}{2 B ^{3}}

Expand

3 (PB + 9 M x + 6 M y) : 3 BP + 27 M x + 18 M y

= \frac{3 BP + 27 M x + 18 M y}{2 B ^{3}}

Apply the fraction rule:

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

\frac{3 BP + 27 M x + 18 M y}{2 B ^{3}} = \frac{3 BP}{2 B ^{3}} + \frac{27 M x}{2 B ^{3}} + \frac{18 M y}{2 B ^{3}}

= \frac{3 BP}{2 B ^{3}} + \frac{27 M x}{2 B ^{3}} + \frac{18 M y}{2 B ^{3}}

Cancel

\frac{3 BP}{2 B ^{3}} : \frac{3 P}{2 B ^{2}}

= \frac{3 P}{2 B ^{2}} + \frac{27 M x}{2 B ^{3}} + \frac{18 M y}{2 B ^{3}}

Divide the numbers:

\frac{18}{2} = 9

= \frac{3 P}{2 B ^{2}} + \frac{27 M x}{2 B ^{3}} + \frac{9 M y}{B ^{3}}

(e^{t} sin (t) - cos (t))^{2}

\frac{d}{d x} ((f (x, y)) (x^{2} + y^{2} - 1))

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

(2 - x) e^{x} - 1

\frac{1}{x ln ( x ) ( 1 + ln ( x ) )}

What is expand (3(PB+9Mx+6My))/(2B^3) ?
The solution to expand (3(PB+9Mx+6My))/(2B^3) is (3P)/(2B^2)+(27Mx)/(2B^3)+(9My)/(B^3)