Question

Please someone answer correctly and fast i will be very grateful to you
Answer no i and iii both

Answer 1

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Answer 2

To Prove:

$i) \: \: \: \: {( {x}^{a + b}) }^{a - b}{( {x}^{b + c}) }^{b - c}{( {x}^{c + a}) }^{c - a} = 1\\ \\ ii) \: \: \: {( \frac{ {x}^{a + b} }{ {x}^{c} } )}^{a + b} {( \frac{ {x}^{b + c} }{ {x}^{a} } )}^{b - c}{( \frac{ {x}^{c + a} }{ {x}^{b} } )}^{c - a} = 1$

Way to Solve:

This question is related to Laws of Indices.

1.No need to Break the exponent in smallest form.

2.All base are same so we don't have to make it hard. So, just follow the Laws of Indices and simplify.

Solution:

See in the Attachment.

${\boxed{\blue{\textsf{Some Important Laws of Indices}}}}$

${a}^{n}.{a}^{m}={a}^{(n + m)}$

${a}^{-1}=\dfrac{1}{a}$

$\dfrac{{a}^{n}}{ {a}^{m}}={a}^{(n-m)}$

${({a}^{c})}^{b}={a}^{b\times c}={a}^{bc}$

${a}^{\frac{1}{x}}=\sqrt[x]{a}\\\\ a^0 = 1$

$[\textsf{Where all variables are real and greater than 0}]$