Question

solve it in sort way its urgent

Answer 1

$= > \frac{ {( {a}^{2} - {b}^{2} ) }^{3} + {( {b}^{2} - {c}^{2}) }^{3} + {( {c}^{2} - {a}^{2}) }^{3} }{(a - b)(b - c)(c - a)(a + b)(b + c)(c + a)} \\ \\ = > \frac{ {( {a}^{2} - {b}^{2} ) }^{3} + {( {b}^{2} - {c}^{2} ) }^{3} + {( {c}^{2} - {a}^{2}) }^{3} }{( {a}^{2} - {b}^{2})( {b}^{2} - {c}^{2} )( {c}^{2} - {a}^{2} )}$
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LET'S :-

=> a² - b² = x

=> b² - c² = y

=> c² - a² = z
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THEN,

$= > \frac{ {x}^{3} + {y}^{3} + {z}^{3} }{x.y.z}$
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here,

=> x + y + z = a²-b²+b²-c²+c²-a²

=> x + y + z = 0 ________[Eq(1)]
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Algebraic Identity :-

$= > {( \alpha + \beta + \gamma )}^{3} = { \alpha }^{3} + { \beta }^{3} + { \gamma }^{3} + ( \alpha + \beta + \gamma )( \alpha \beta + \beta \gamma + \gamma \alpha ) \\ \\ = > { \alpha }^{3} + { \beta }^{3} + { \gamma }^{3} = {( \alpha + \beta + \gamma )}^{3} - ( \alpha + \beta + \gamma )( \alpha \beta + \beta \gamma + \gamma \alpha )$
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THEN,

$= > \frac{ {x}^{3} + {y}^{3} + {z}^{3} }{x.y.z} \\ \\ = > \frac{ {(x + y + z)}^{3} - (x + y + z)(xy + yz + zx)}{x.y.z} \\ \\ = > \frac{ {0}^{3} - 0 \times (xy + yz + zx)}{xyz} \\ \\ = > 0$
____________[ANSWER]

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_-_-_-_☆$BE \: \: \: \: BRAINLY$☆_-_-_-_

Answer 2

Hey friend, Harish here.

Please refer to the image attached for the answer.

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Hope my answer is helpful to you.