Question
Simplify the following expressions: $\big(\text{x}+\text{y}+\text{z}\big)^2+\Big(\text{x}+\frac{\text{y}}{2}+\frac{\text{z}}{3}\Big)^2-\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}+\frac{\text{z}}{4}\Big)^2$
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Answer
Expanding, we get $zx=\big[\text{x}^2+\text{y}^2+\text{z}^2+2\text{xy}+2\text{yz}+2\text{}\big]\\+\Big[\text{x}^2+\frac{\text{y}^2}{4}+\frac{\text{z}^2}{9}+2\text{x}\frac{\text{y}}{2}+2\frac{\text{zx}}{3}+\frac{\text{yz}}{3}\Big]\\-\Big[\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}+\frac{\text{z}^2}{10}+\frac{\text{xy}}{3}+\frac{\text{yz}}{6}+\frac{\text{xz}}{4}\Big]$ $\big[\therefore\big(\text{x}+\text{y}+\text{z}\big)^2=\text{x}^2+\text{y}^2+\text{z}^2+2\text{xy}+2\text{yz}+2\text{zx}\big]$ $=\text{x}^2+\text{y}^2+\text{z}^2+2\text{xy}+2\text{yz}+2\text{zx}+\text{x}^2+\frac{\text{y}^2}{4}+\frac{\text{z}^2}{9}\\+2\text{x}\frac{\text{y}}{2}+\frac{\text{xy}}{3}+\frac{2\text{zx}}{3}-\frac{\text{x}^2}{4}-\frac{\text{y}^2}{9}-\frac{\text{z}^2}{10}-\frac{\text{xy}}{3}-\frac{\text{yz}}{6}-\frac{\text{xz}}{4}$ Rearranging coefficients, $=\frac{8\text{x}^2-\text{x}^2}{4}+\frac{36\text{y}^2+9\text{y}^2-4\text{y}^2}{36}+\frac{144\text{z}^2+16\text{z}^2-9\text{z}^2}{144}\\+\frac{6\text{xy}+3\text{xy}-\text{xy}}{3}\frac{13\text{yz}}{6}+\frac{29\text{xz}}{12}$ $=\frac{7\text{x}^2}{4}+\frac{41\text{y}^2}{36}+\frac{151\text{z}^2}{144}+\frac{8\text{xy}}{3}+\frac{13\text{yz}}{6}+\frac{29\text{zx}}{12}$ $\big(\text{x}+\text{y}+\text{z}\big)^2+\Big(\text{x}+\frac{\text{y}}{3}+\frac{\text{z}}{3}\Big)^2-\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}+\frac{\text{z}}{4}\Big)^2$ $=\frac{7\text{x}^2}{4}+\frac{41\text{y}^2}{36}+\frac{151\text{z}^2}{144}+\frac{8\text{xy}}{3}+\frac{13\text{yz}}{6}+\frac{29\text{zx}}{12}$
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