What is expand ((a^2b^3c-4)2)/((abc)3) ?

The solution to expand ((a^2b^3c-4)2)/((abc)3) is (2ab^2)/3-8/(3abc)

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expand ((a^2b^3c-4)2)/((abc)3)

expand

\frac{( a ^{2} b ^{3} c - 4 ) 2}{( ab c ) 3}

\frac{2 a b ^{2}}{3} - \frac{8}{3 ab c}

Solution steps

\frac{( a ^{2} b ^{3} c - 4 ) \cdot 2}{( ab c ) \cdot 3}

Remove parentheses:

(a) = a

= \frac{( a ^{2} b ^{3} c - 4 ) \cdot 2}{ab c \cdot 3}

Expand

(a^{2} b^{3} c - 4) \cdot 2 : 2 a^{2} b^{3} c - 8

= \frac{2 a ^{2} b ^{3} c - 8}{3 ab c}

Apply the fraction rule:

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

\frac{2 a ^{2} b ^{3} c - 8}{ab c \cdot 3} = \frac{2 a ^{2} b ^{3} c}{ab c \cdot 3} - \frac{8}{ab c \cdot 3}

= \frac{2 a ^{2} b ^{3} c}{3 ab c} - \frac{8}{3 ab c}

Cancel

\frac{2 a ^{2} b ^{3} c}{ab c \cdot 3} : \frac{2 a b ^{2}}{3}

= \frac{2 a b ^{2}}{3} - \frac{8}{3 ab c}

\frac{6 x}{( x ^{2} + 2 ) ^{2}}

\frac{( sec ( x ) ) ^{2} β - sin ( β ) tan ( β )}{cos ( x )}

- \frac{1}{s ( s + 2 )}

x - (2 x + 1 = 8 - (3 x + 3))

\frac{s}{( s - 2 ) ^{2}}

What is expand ((a^2b^3c-4)2)/((abc)3) ?
The solution to expand ((a^2b^3c-4)2)/((abc)3) is (2ab^2)/3-8/(3abc)