What is expand ((8b+9)^3)/(1728) ?

The solution to expand ((8b+9)^3)/(1728) is (8b^3}{27}+b^2+\frac{9b)/8+27/64

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expand ((8b+9)^3)/(1728)

expand

\frac{( 8 b + 9 ) ^{3}}{1728}

\frac{8 b ^{3}}{27} + b^{2} + \frac{9 b}{8} + \frac{27}{64}

Solution steps

\frac{( 8 b + 9 ) ^{3}}{1728}

(8 b + 9)^{3} = 512 b^{3} + 1728 b^{2} + 1944 b + 729

= \frac{512 b ^{3} + 1728 b ^{2} + 1944 b + 729}{1728}

Apply the fraction rule:

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

\frac{512 b ^{3} + 1728 b ^{2} + 1944 b + 729}{1728} = \frac{512 b ^{3}}{1728} + \frac{1728 b ^{2}}{1728} + \frac{1944 b}{1728} + \frac{729}{1728}

= \frac{512 b ^{3}}{1728} + \frac{1728 b ^{2}}{1728} + \frac{1944 b}{1728} + \frac{729}{1728}

\frac{512 b ^{3}}{1728} = \frac{8 b ^{3}}{27}

\frac{1728 b ^{2}}{1728} = b^{2}

\frac{1944 b}{1728} = \frac{9 b}{8}

\frac{729}{1728} = \frac{27}{64}

= \frac{8 b ^{3}}{27} + b^{2} + \frac{9 b}{8} + \frac{27}{64}

(4 x^{2} + x^{2} u - 3 u^{2} x^{2}) d x

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

\frac{1}{x ^{2} ( x ^{2} + 25 )}

\frac{( x - 3 ) ( x - 2 )}{2 ( x + 3 )}

, (9 + 3 i) (9 - 3 i)

What is expand ((8b+9)^3)/(1728) ?
The solution to expand ((8b+9)^3)/(1728) is (8b^3}{27}+b^2+\frac{9b)/8+27/64