Question

Prove that....................​

Answer 1

Required Answer:-

Given to prove:

$\rm \mapsto \sqrt{ {x}^{ - 1}y} \times \sqrt{ {y}^{ - 1}z } \times \sqrt{ {z}^{ - 1} x} = 1$

Proof:

Taking LHS,

$\rm \sqrt{ {x}^{ - 1}y} \times \sqrt{ {y}^{ - 1}z } \times \sqrt{ {z}^{ - 1} x}$

$\rm = \sqrt{ {x}^{ - 1}y\times {y}^{ - 1}z \times {z}^{ - 1} x}$

$\rm = \sqrt{ \dfrac{ \cancel{y}}{ \cancel{x} }\times \dfrac{ \cancel{z}}{ \cancel{y}} \times \dfrac{ \cancel{x}}{ \cancel{z} }}$

$\rm = 1$

= RHS (Hence Proved)

Important formula to solve indices questions:

$\rm \mapsto {x}^{a} \times {x}^{b} = {x}^{a + b}$

$\rm \mapsto ({x}^{a})^{b} = {x}^{ab}$

$\rm \mapsto \dfrac{ {x}^{a} }{ {x}^{b} } = {x}^{a - b}$

$\rm \mapsto {x}^{0} = 1 \: \: (x \neq 0)$

$\rm \mapsto {x}^{y} = \dfrac{1}{ {x}^{ - y} }$

$\rm \mapsto {x}^{a} = {y}^{a} \implies x = y \: \: (a \neq 0)$

$\rm \mapsto \sqrt[x]{y} = {y}^{ \dfrac{1}{x} }$

$\rm \mapsto \dfrac{ {x}^{a} }{ {y}^{a} } = \bigg(\dfrac{x}{y} \bigg)^{a}$

Answer 2

$\sf{\sqrt{x^{-1}y}\sqrt{y^{-1}z}\sqrt{z^{-1}x}}$

$\sf{=\sqrt{x^{-1}xy^{-1}yz^{-1}z}}$

$\sf{=\sqrt{\dfrac{x}{x}\cdot \dfrac{y}{y}\cdot \dfrac{z}{z}}}$

$\sf{=\sqrt{1\cdot1\cdot1}}$

$\sf{=\sqrt{1}}$

$\sf{=1}$