Question

The factorized form of $\frac{1}{xyz}\left(x^2+y^2+z^2\right)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$ is _____________.

Answer 1

$$\begin{gathered} \frac{1}{xyz}\left(x^2+y^2+z^2\right)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \\ =\frac{\left(x^2+y^2+z^2\right)}{xyz}+2\left(\frac{yz+xz+xy}{xyx}\right) \\ =\frac{x^2+y^2+z^2+2\left(yz+xz+xy\right)}{xyz} \\ =\frac{{\left(x+y+z\right)}^2}{xyz}\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:{\left(a+b+c\right)}^2=a^2+b^2+c^2+2ab+2bc+2ac\right) \\ =\frac{1}{xyz}{\left(x+y+z\right)}^2 \end{gathered}$$

Hence, the factorized form of $\frac{1}{xyz}\left(x^2+y^2+z^2\right)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$ is $\underline{\frac{1}{xyz}{\left(x+y+z\right)}^2}.$