Question

To prove that........................​

Answer 1

Required Answer:-

Given to prove:

$\rm \mapsto \bigg(\dfrac{ {x}^{a} }{ {x}^{b} } \bigg)^{ {a}^{2} + ab + {b}^{2} } \times { \bigg( \dfrac{ {x}^{b} }{ {x}^{c} } \bigg)}^{ {b}^{2} + bc + {c}^{2} } \times { \bigg( \dfrac{ {x}^{c} }{ {x}^{a} } \bigg)}^{ {c}^{2} + ac + {a}^{2} } = 1$

Proof:

Taking LHS,

$\rm \bigg(\dfrac{ {x}^{a} }{ {x}^{b} } \bigg)^{ {a}^{2} + ab + {b}^{2} } \times { \bigg( \dfrac{ {x}^{b} }{ {x}^{c} } \bigg)}^{ {b}^{2} + bc + {c}^{2} } \times { \bigg( \dfrac{ {x}^{c} }{ {x}^{a} } \bigg)}^{ {c}^{2} + ac + {a}^{2} }$

$\rm = {x}^{(a - b)( {a}^{2} + ab + {b}^{2}) } \times {x}^{(b - c)({b}^{2} + bc + {c}^{2} ) } \times {x}^{(c - a)( {c}^{2} + ac + {a}^{2})}$

$\rm = {x}^{ ({a}^{3} - {b}^{3} )} \times {x}^{( {b}^{3} - {c}^{3}) } \times {x}^{ ({c}^{3} - {a}^{3} )}$

$\rm = {x}^{ {a}^{3} - {b}^{3} + {b}^{3} - {c}^{3} + {c}^{3} - {a}^{3} }$

$\rm = {x}^{0 }$

$\rm = 1$

= 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

Answer:

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