Question
Evaluate: $\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3=\Big(\frac{5}{6}\Big)^3$
✓
Answer
Let $\text{a}=\frac{1}{2},\ \text{b}=\frac{1}{3}$ and $\text{c}=\frac{-5}{6}$ Then,
$\text{a} + \text{b} + \text{c}=\frac{1}{2}+\frac{1}{3}-\frac{5}{6}$
$=\frac{3+2}{6}-\frac{5}{6}$
$\Rightarrow\text{a}+\text{b}+\text{c}=\frac{5}{6}-\frac{5}{6}=0$
$\therefore\text{a}^3+\text{b}^3+\text{c}^3=3\text{abc}$
$\Rightarrow\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3+\Big(\frac{-5}{6}\Big)^3=3\times\Big(\frac{1}{2}\Big)\times\Big(\frac{1}{3}\Big)\times\Big(\frac{-5}{6}\Big)$
$=\frac{-5}{12}$
$\therefore\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3-\Big(\frac{5}{6}\Big)^3=\frac{-5}{12}$
$\text{a} + \text{b} + \text{c}=\frac{1}{2}+\frac{1}{3}-\frac{5}{6}$
$=\frac{3+2}{6}-\frac{5}{6}$
$\Rightarrow\text{a}+\text{b}+\text{c}=\frac{5}{6}-\frac{5}{6}=0$
$\therefore\text{a}^3+\text{b}^3+\text{c}^3=3\text{abc}$
$\Rightarrow\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3+\Big(\frac{-5}{6}\Big)^3=3\times\Big(\frac{1}{2}\Big)\times\Big(\frac{1}{3}\Big)\times\Big(\frac{-5}{6}\Big)$
$=\frac{-5}{12}$
$\therefore\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3-\Big(\frac{5}{6}\Big)^3=\frac{-5}{12}$
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