Question

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If a+b+c = 0, then $x^{a^{2}b^{-1}c^{-1}} . x^{a^{-1}b^{2}c^{-1}} . x^{a^{-1}b^{-1}c^{2}}$ has its value as :

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Answer 1

Hey Friend ☺

The above expression is written as x^3

Your required steps are in above attachment refer it

Hope it helps you ..!!

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Answer 2

The answer is given below :

Now,

${a}^{2} {b}^{ - 1} {c}^{ - 1} + {a}^{ - 1} {b}^{2} {c}^{ - 1} + {a}^{ - 1} {b}^{ - 1} {c}^{2} \\ \\ = \frac{ {a}^{2} }{bc} + \frac{ {b}^{2} }{ca} + \frac{ {c}^{2} }{ab} \\ \\ = \frac{ {a}^{3} + {b}^{3} + {c}^{3} }{abc} \\ \\ = \frac{( a+b + c) ( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) + 3abc }{abc} \\ \\ = \frac{0 + 3abc}{abc} \: \: since \: \: a + b + c = 0 \\ \\ = \frac{3abc}{abc} \\ \\ = 3$


So,

${x}^{ {a}^{2} {b}^{ - 1} {c}^{ - 1} }. {x}^{ {a}^{ - 1} {b}^{ 2} {c}^{ - 1} }. {x}^{ {a}^{ - 1} {b}^{ - 1} {c}^{ 2} } \\ \\ = {x}^{( {a}^{2} {b}^{ - 1} {c}^{ - 1} + {a}^{ - 1} {b}^{2} {c}^{ - 1} + {a}^{ - 1} {b}^{ - 1} {c}^{2} )} \\ \\ = {x}^{3} \: \: by \: \: the \: \: previous \: \: calculation$

which is the required value.

Thank you for your question.