Question
Expand:
$\Big(3\text{a}+\frac{1}{4\text{b}}\Big)^3$
$\Big(3\text{a}+\frac{1}{4\text{b}}\Big)^3$
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Answer
$\Big(3\text{a}+\frac{1}{4\text{b}}\Big)^3$
$=(3\text{a})^3+\Big(\frac{1}{4\text{b}}\Big)^3+3(3\text{a})^2\Big(\frac{1}{4\text{b}}\Big)+3(3\text{a})\Big(\frac{1}{4\text{b}}\Big)^2$
$=27\text{a}^3+\frac{1}{64\text{b}^3}+\frac{27\text{a}^2}{4\text{b}}+\frac{9\text{a}}{16\text{b}^2}$
$=(3\text{a})^3+\Big(\frac{1}{4\text{b}}\Big)^3+3(3\text{a})^2\Big(\frac{1}{4\text{b}}\Big)+3(3\text{a})\Big(\frac{1}{4\text{b}}\Big)^2$
$=27\text{a}^3+\frac{1}{64\text{b}^3}+\frac{27\text{a}^2}{4\text{b}}+\frac{9\text{a}}{16\text{b}^2}$
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