Question

teeno teeno ke teeno bata do bhaiyo or bheno pllz bhai help kl paper hai ​

Answer 1

1.

$\frac{ {x}^{2a + 2b} {x}^{2b + 2c} {x}^{2c + 2a} }{ {x}^{4a + 4b + 4c} } \\ if \: bases \: are \: equal \: powers \:should \: be \: added \\ = > \frac{ {x}^{2a + 2b + 2b + 2c + 2c + 2a} }{{x}^{4a + 4b + 4c}} \\ = > \frac{{x}^{4a + 4b + 4c}}{{x}^{4a + 4b + 4c}} \\ = > 1$

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2.

${ (\frac{ {x}^{ {a}^{2} } }{ {x}^{ {b}^{2} } } )}^{ \frac{1}{a + b} } \: { (\frac{ {x}^{ {b}^{2} } }{ {x}^{ {c}^{2} } } )}^{ \frac{1}{b + c} } \: { (\frac{ {x}^{ {c}^{2} } }{ {x}^{ {c}^{2} } } )}^{ \frac{1}{c + a} } \\ = > {( {x}^{ {a}^{2} - {b}^{2} } )}^{ \frac{1}{a + b} } \: {( {x}^{ {b}^{2} - {c}^{2} } )}^{ \frac{1}{b + c} } \: {( {x}^{ {c}^{2} - {a}^{2} } )}^{ \frac{1}{c + a} } \\ = > {x}^{ \frac{ {a}^{2} - {b}^{2} }{a + b} } \: {x}^{ \frac{ {b}^{2} - {c}^{2} }{b + c} } \: {x}^{ \frac{ {c}^{2} - {a}^{2} }{c + a} } \\ = > {x}^{a - b} {x}^{b - c} {x}^{c - a} \\ = > {x}^{a - b + b - c + c - a} \\ = > {x}^{0} = 1$

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3.

this question is wrong becouse it's not maintaining any 3 proper sets

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