Question

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Answer 1

Step-by-step explanation:

$( {x}^{(a - b)} )^{a + b} .( {x}^{(b - c)} )^{b + c}( {x}^{(c - a)} )^{c + a} = 1 \\ LHS = ( {x}^{(a - b)} )^{a + b} .( {x}^{(b - c)} )^{b + c}( {x}^{(c - a)} )^{c + a} \\ = {x}^{ {a}^{2} - {b}^{2} } .{x}^{ {b}^{2} - {c}^{2} } .{x} ^{ {c}^{2} - {a}^{2} } \\ {x}^{{a}^{2} - {b}^{2} + {b}^{2} - {c}^{2} +{c}^{2} - {a}^{2} } \\ = x ^{0} \\ = 1 \\ = RHS \\ thus \: proved$