What is expand ((2a^2b^4-a^4b)^3)/8 ?

The solution to expand ((2a^2b^4-a^4b)^3)/8 is a^6b^{12}-(3a^8b^9)/2+(3a^{10}b^6)/4-(a^{12}b^3)/8

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expand ((2a^2b^4-a^4b)^3)/8

expand

\frac{( 2 a ^{2} b ^{4} - a ^{4} b ) ^{3}}{8}

a^{6} b^{12} - \frac{3 a ^{8} b ^{9}}{2} + \frac{3 a ^{10} b ^{6}}{4} - \frac{a ^{12} b ^{3}}{8}

Solution steps

\frac{( 2 a ^{2} b ^{4} - a ^{4} b ) ^{3}}{8}

(2 a^{2} b^{4} - a^{4} b)^{3} = 8 a^{6} b^{12} - 12 a^{8} b^{9} + 6 a^{10} b^{6} - a^{12} b^{3}

= \frac{8 a ^{6} b ^{12} - 12 a ^{8} b ^{9} + 6 a ^{10} b ^{6} - a ^{12} b ^{3}}{8}

Apply the fraction rule:

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

\frac{8 a ^{6} b ^{12} - 12 a ^{8} b ^{9} + 6 a ^{10} b ^{6} - a ^{12} b ^{3}}{8} = \frac{8 a ^{6} b ^{12}}{8} - \frac{12 a ^{8} b ^{9}}{8} + \frac{6 a ^{10} b ^{6}}{8} - \frac{a ^{12} b ^{3}}{8}

= \frac{8 a ^{6} b ^{12}}{8} - \frac{12 a ^{8} b ^{9}}{8} + \frac{6 a ^{10} b ^{6}}{8} - \frac{a ^{12} b ^{3}}{8}

Divide the numbers:

\frac{8}{8} = 1

= a^{6} b^{12} - \frac{12 a ^{8} b ^{9}}{8} + \frac{6 a ^{10} b ^{6}}{8} - \frac{a ^{12} b ^{3}}{8}

Cancel

\frac{12 a ^{8} b ^{9}}{8} : \frac{3 a ^{8} b ^{9}}{2}

= a^{6} b^{12} - \frac{3 a ^{8} b ^{9}}{2} + \frac{6 a ^{10} b ^{6}}{8} - \frac{a ^{12} b ^{3}}{8}

Cancel

\frac{6 a ^{10} b ^{6}}{8} : \frac{3 a ^{10} b ^{6}}{4}

= a^{6} b^{12} - \frac{3 a ^{8} b ^{9}}{2} + \frac{3 a ^{10} b ^{6}}{4} - \frac{a ^{12} b ^{3}}{8}

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

(5 b, 3, - c) (6, a, 2)

e^{x} (x + x^{2})

5^{2} - 14 \cdot 5 + 54

2 (- \frac{ln ( x )}{3 x ^{3}} - \frac{1}{9 x ^{3}}) + C

What is expand ((2a^2b^4-a^4b)^3)/8 ?
The solution to expand ((2a^2b^4-a^4b)^3)/8 is a^6b^{12}-(3a^8b^9)/2+(3a^{10}b^6)/4-(a^{12}b^3)/8