Question
Factorize the following expressions:$\Big[\frac{\text{x}}{2}+\text{y}+\frac{\text{z}}{3}\Big]^3+\Big[\frac{\text{x}}{3}-\frac{2\text{y}}{3}+\text{z}\Big]^3+\Big[-\frac{5\text{x}}{6}-\frac{\text{y}}{3}-\frac{4\text{z}}{3}\Big]^3$
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Answer
$\Big[\frac{\text{x}}{2}+\text{y}+\frac{\text{z}}{3}\Big]^3+\Big[\frac{\text{x}}{3}-\frac{2\text{y}}{3}+\text{z}\Big]^3+\Big[-\frac{5\text{x}}{6}-\frac{\text{y}}{3}-\frac{4\text{z}}{3}\Big]^3$Let $\Big(\frac{\text{x}}{2}+\text{y}+\frac{\text{z}}{3}\Big)=\text{a},\Big(\frac{\text{x}}{3}-\frac{2\text{y}}{3}+\text{z}\Big)=\text{b},\Big(-\frac{5\text{x}}{6}-\frac{\text{y}}{3}-\frac{4\text{z}}{3}\Big)=\text{c}$
$\text{a}+\text{b}+\text{c}=\frac{\text{x}}{2}+\text{y}+\frac{\text{z}}{3}+\frac{\text{x}}{3}-\frac{\text{2y}}{3}+\text{z}-\frac{5\text{x}}{6}-\frac{\text{y}}{3}-\frac{4\text{z}}{3}$
$\text{a}+\text{b}+\text{c}=\Big(\frac{\text{x}}{2}+\frac{\text{x}}{3}-\frac{5\text{x}}{6}\Big)+\Big(\text{y}-\frac{\text{2y}}{3}-\frac{\text{y}}{3}\Big)+\Big(\frac{\text{z}}{3}+\text{z}-\frac{4\text{z}}{3}\Big)$
$\text{a}+\text{b}+\text{c}=\frac{3\text{x}}{6}+\frac{2\text{x}}{6}-\frac{5\text{x}}{6}+\frac{3\text{y}}{3}-\frac{2\text{y}}{3}-\frac{\text{y}}{3}+\frac{\text{z}}{3}+\frac{3\text{z}}{3}-\frac{4\text{z}}{3}$
$\text{a}+\text{b}+\text{c}=\frac{5\text{x}-5\text{x}}{6}+\frac{3\text{y}-3\text{y}}{3}+\frac{4\text{z}-4\text{z}}{3}$
$\text{a}+\text{b}+\text{c}=0$
$\because \ \text{a}+\text{b}+\text{c}=0$
$\therefore \ \text{a}^3+\text{b}^3+\text{c}^3=3\text{abc}$
$\therefore\Big[\frac{\text{x}}{2}+\text{y}+\frac{\text{z}}{3}\Big]^3+\Big[\frac{\text{x}}{3}-\frac{2\text{y}}{3}+\text{z}\Big]^3+\Big[-\frac{5\text{x}}{6}-\frac{\text{y}}{3}-\frac{4\text{z}}{3}\Big]^3$
$=3\Big(\frac{\text{x}}{2}+\text{y}+\frac{\text{z}}{3}\Big)\Big(\frac{\text{x}}{3}-\frac{2\text{y}}{3}+\text{z}\Big)\Big(-\frac{5\text{x}}{6}-\frac{\text{y}}{3}-\frac{4\text{z}}{3}\Big)$
$\text{a}+\text{b}+\text{c}=\frac{\text{x}}{2}+\text{y}+\frac{\text{z}}{3}+\frac{\text{x}}{3}-\frac{\text{2y}}{3}+\text{z}-\frac{5\text{x}}{6}-\frac{\text{y}}{3}-\frac{4\text{z}}{3}$
$\text{a}+\text{b}+\text{c}=\Big(\frac{\text{x}}{2}+\frac{\text{x}}{3}-\frac{5\text{x}}{6}\Big)+\Big(\text{y}-\frac{\text{2y}}{3}-\frac{\text{y}}{3}\Big)+\Big(\frac{\text{z}}{3}+\text{z}-\frac{4\text{z}}{3}\Big)$
$\text{a}+\text{b}+\text{c}=\frac{3\text{x}}{6}+\frac{2\text{x}}{6}-\frac{5\text{x}}{6}+\frac{3\text{y}}{3}-\frac{2\text{y}}{3}-\frac{\text{y}}{3}+\frac{\text{z}}{3}+\frac{3\text{z}}{3}-\frac{4\text{z}}{3}$
$\text{a}+\text{b}+\text{c}=\frac{5\text{x}-5\text{x}}{6}+\frac{3\text{y}-3\text{y}}{3}+\frac{4\text{z}-4\text{z}}{3}$
$\text{a}+\text{b}+\text{c}=0$
$\because \ \text{a}+\text{b}+\text{c}=0$
$\therefore \ \text{a}^3+\text{b}^3+\text{c}^3=3\text{abc}$
$\therefore\Big[\frac{\text{x}}{2}+\text{y}+\frac{\text{z}}{3}\Big]^3+\Big[\frac{\text{x}}{3}-\frac{2\text{y}}{3}+\text{z}\Big]^3+\Big[-\frac{5\text{x}}{6}-\frac{\text{y}}{3}-\frac{4\text{z}}{3}\Big]^3$
$=3\Big(\frac{\text{x}}{2}+\text{y}+\frac{\text{z}}{3}\Big)\Big(\frac{\text{x}}{3}-\frac{2\text{y}}{3}+\text{z}\Big)\Big(-\frac{5\text{x}}{6}-\frac{\text{y}}{3}-\frac{4\text{z}}{3}\Big)$
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