Question

How to solve the above maths question? With step-by-step method

Answer 1

Hey friend,
Here is the answer you were looking for:
${( \frac{ {x}^{a} }{ {x}^{b} }) }^{a + b} \times {( \frac{ {x}^{b} }{ {x}^{c} } )}^{b + c} \times {( \frac{ {x}^{c} }{ {x}^{a} }) }^{c - a} = 1$

Using the identity :

$\frac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$

Taking L.H.S.

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

Using the identity :

$(a + b)(a - b) = {a}^{2} - {b}^{2}$

$= {x}^{ {a}^{2} - {b}^{2} } \times {x}^{ {b}^{2} - {c}^{2} } \times {x}^{ {c}^{2} - {a}^{2} }$


Using the identity :

${a}^{m} \times {a}^{n} = {a}^{m + n}$


$= {x}^{ {a}^{2} - {b}^{2} + {b}^{2} - {c}^{2} + {c}^{2} - {a}^{2} } \\ \\ = {x}^{0} \\ \\ 1 = 1$

LHS = RHS

Hence proved.

Hope this helps!!

If you have any doubt regarding to my answer, feel free to ask in the comment section or inbox me if needed.

@Mahak24

Thanks...
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