Question

Prove that, [ (x)^(a²) ÷ (x)^(b²) ]^{1/(a+b)} × [ (x)^(b²) ÷ (x)^(c²) ]^{1/(b+c)} × [ (x)^(c²) ÷ (x)^(a²) ]^{1/(c+a)} = 1

Answer 1

SOLUTION IS IN THE ATTACHED IMAGE.


LHS = $\bigg [ \frac{x ^{ {a}^{2} } }{ {x}^{ {b}^{2} } } \bigg] ^{ \frac{1}{a + b} }$ × $\bigg [ \frac{x ^{ {b}^{2} } }{ {x}^{ {c}^{2} } } \bigg] ^{ \frac{1}{b + c} }$ × $\bigg [ \frac{x ^{ {c}^{2} } }{ {x}^{ {a}^{2} } } \bigg] ^{ \frac{1}{c + a} }$

.

= ${ \bigg[ (x)^{ a^{2} - b^{2} } \bigg] }^{\frac{1}{a+b}}$×${ \bigg[ (x)^{ b^{2} - c^{2} } \bigg] }^{\frac{1}{b + c}}$×${ \bigg[ (x)^{ c^{2} - a^{2} } \bigg] }^{\frac{1}{c + a}}$

.

[ $\frac {a^{m}}{a^{n}}\: =\: a^{m-n}$ ]

.

= $[(x)^{(a-b) (a+b)}]^{\frac{1}{a+b}}$ × $[(x)^{(b-c) (b+c)}]^{\frac{1}{b+c}}$ × $[(x)^{(c-a) (c+a)}]^{\frac{1}{c+a}}$

.

[ $\therefore$ $\ {a}^{2} - {b}^{2}}$ = (a-b) (a+b) ]

.

= $\ {(x)}^{a-b}$ × $\ {(x)}^{b-c}$ × $\ {(x)}^{c-a}$

.

[ $\therefore (a^{m})^{n} = a^{mn}$ ]

.

= $\ {(x)}^{a-b+b-c+c-a}$

.

[ $\therefore {a}^{m} \times {a}^{n} = {a}^{m+n}$ ]

.

= $\ {x}^{0}$ = 1

Hence proved


Hope it helps

Answer 2

CHECK OUT THE ATTACHMENT
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