Question

The square root of$\frac{1}{abc}(a^2{b}^2{c}^2+2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})is(a,b,c\ne 0))$

• A. $\frac{\sqrt{a+b+c}}{abc}$
• B. $\sqrt{\frac{a}{bc}}+\sqrt{\frac{b}{ca}}+\sqrt{\frac{c}{ab}}$
• C. $\sqrt{\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{a}}$
• D. $\sqrt{abc}$

Answer 1

The correct option is A $\frac{\sqrt{a+b+c}}{abc}$

So,

$\begin{array}{c}\sqrt{\frac{1}{abc}\left(a^2{b}^2{c}^2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right)}\\ =\sqrt{\frac{1}{abc}\left(\frac{a^3{b}^3{c}^3+2bc+2ac+2ab}{abc}\right)}\\ =\sqrt{\frac{a^3{b}^3{c}^3+2bc+2ac+2ab}{a^2{b}^2{c}^2}}\\ Hence,\\ \frac{\sqrt{a+b+c}}{abc}\end{array}$

Option $\left(A\right)$ is correct answer