Question

Square root of $\dfrac{1}{{xyz}}\left( {{x^2} + {y^2} + {z^2}} \right) + 2\left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right)$ is,
A) $\sqrt {\dfrac{x}{{yz}}} + \sqrt {\dfrac{y}{{zx}}} + \sqrt {\dfrac{z}{{xy}}} $
B) $\sqrt {\dfrac{{yz}}{x}} + \sqrt {\dfrac{{zx}}{y}} + \sqrt {\dfrac{{xy}}{z}} $
C) ${\left( {\dfrac{x}{{yz}}} \right)^2} + {\left( {\dfrac{y}{{zx}}} \right)^2} + {\left( {\dfrac{z}{{xy}}} \right)^2}$
D) ${\left( {\dfrac{{yz}}{x}} \right)^{\dfrac{1}{2}}} + {\left( {\dfrac{{zx}}{y}} \right)^{\dfrac{1}{2}}} + {\left( {\dfrac{{xy}}{z}} \right)^{\dfrac{1}{2}}}$

Answer 1

Hint: First, take LCM in the second term. After that take $\dfrac{1}{{xyz}}$ common from both terms. Then use the formula ${\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right)$ to make the square of sum of x, y, and z. After that take the square root of the terms. Then multiply each variable by the common and cancel out the common factors to get the desired result.

Complete step-by-step solution:
Given:- $\dfrac{1}{{xyz}}\left( {{x^2} + {y^2} + {z^2}} \right) + 2\left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right)$
Take LCM in the second terms,
$ \Rightarrow \dfrac{1}{{xyz}}\left( {{x^2} + {y^2} + {z^2}} \right) + 2\left( {\dfrac{{yz + zx + xy}}{{xyz}}} \right)$
Now take $\dfrac{1}{{xyz}}$ common from both terms,
$ \Rightarrow \dfrac{1}{{xyz}}\left[ {{x^2} + {y^2} + {z^2} + 2\left( {xy + yz + zx} \right)} \right]$
We know that,
${\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right)$
Use the formula in the above expression,
$ \Rightarrow \dfrac{1}{{xyz}}{\left( {x + y + z} \right)^2}$
Take square root on the expression,
$ \Rightarrow \sqrt {\dfrac{1}{{xyz}}{{\left( {x + y + z} \right)}^2}} $
Simplify the expression,
$ \Rightarrow \dfrac{1}{{\sqrt {xyz} }}\sqrt {{{\left( {x + y + z} \right)}^2}} $
Cancel out root with the power,
$ \Rightarrow \dfrac{1}{{\sqrt {xyz} }}\left( {x + y + z} \right)$
Multiply the common factor with each term,
\[ \Rightarrow \dfrac{x}{{\sqrt {xyz} }} + \dfrac{y}{{\sqrt {xyz} }} + \dfrac{z}{{\sqrt {xyz} }}\]
Use the identity $x = {\left( {\sqrt x } \right)^2}$,
\[ \Rightarrow \dfrac{{{{\left( {\sqrt x } \right)}^2}}}{{\sqrt {xyz} }} + \dfrac{{{{\left( {\sqrt y } \right)}^2}}}{{\sqrt {xyz} }} + \dfrac{{{{\left( {\sqrt z } \right)}^2}}}{{\sqrt {xyz} }}\]
Cancel out the common factors from each term,
\[ \Rightarrow \dfrac{{\sqrt x }}{{\sqrt {yz} }} + \dfrac{{\sqrt y }}{{\sqrt {xz} }} + \dfrac{{\sqrt z }}{{\sqrt {xy} }}\]
We know that,
$\dfrac{{\sqrt a }}{{\sqrt b }} = \sqrt {\dfrac{a}{b}} $
Use this in the above expression,
$\therefore \sqrt {\dfrac{x}{{yz}}} + \sqrt {\dfrac{y}{{zx}}} + \sqrt {\dfrac{z}{{xy}}} $
Thus, the square root is $\sqrt {\dfrac{x}{{yz}}} + \sqrt {\dfrac{y}{{zx}}} + \sqrt {\dfrac{z}{{xy}}} $.

Hence, option (A) is the correct answer.

Note: The students may start solving this question the wrong way by simply multiplying the common factor in the brackets. It will lead to either lengthy calculation or give the wrong result.
The square root of a number is a value, which on multiplied by itself gives the original number. Suppose, x is the square root of y, then it is represented as \[x = \sqrt y \] or we can express the same equation as ${x^2} = y$. Here, ’$\sqrt {} $’ is the radical symbol used to represent the root of numbers. The positive number, when multiplied by itself, represents the square of the number. The square root of the square of a positive number gives the original number.