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Question 1. Asked on :16 January 2019:05:34:22 PM

#### (a2-b2)3+(b2-c2)3+(c2-a2)3(a-b)3+(b-c)3+(c-a)3Simplify this Equation:

-Added by Shalu Shukla Mathematics » Ch-2

Nitish kumar

( a² - b² )³ + ( b² - c² )³ + ( c² - a² )³ ] / [ ( a - b )³ + ( b - c )³ + ( c - a )³ ] == 3( a² - b² )( b² - c² )( c² - a² ) / 3( a - b )( b - c )( c - a ) ... In triangle abc, p is the midpoint of seg ab.

-Answered by Nitish kumar On 17 January 2019:10:36:38 AM(740Average Rating Based on rating)

Himanshi Verma

We know,

---> if a + b + c = 0 ; a³ + b³ + c³ = 3abc

We observe,

( a² - b² ) + ( b² - c² ) + ( c² - a² ) = 0

=> ( a² - b² )³ + ( b² - c² )³ + ( c² - a² )³ = 3( a² - b² )( b² - c² )( c² - a² )

Similarly,

( a - b ) + ( b - c ) + ( c - a ) = 0

=> ( a - b )³ + ( b - c )³ + ( c - a )³ = 3( a - b )( b - c )( c - a )

Now, [ ( a² - b² )³ + ( b² - c² )³ + ( c² - a² )³ ] / [ ( a - b )³ + ( b - c )³ + ( c - a )³ ]

== 3( a² - b² )( b² - c² )( c² - a² ) / 3( a - b )( b - c )( c - a )

= ( a + b )( b + c )( c + a ) <---- Answer..

-Answered by Himanshi Verma On 03 February 2019:04:45:43 PM(1270Average Rating Based on rating)

Master Purushottam

We know,

---> if a + b + c = 0 ; a³ + b³ + c³ = 3abc

We observe,

( a² - b² ) + ( b² - c² ) + ( c² - a² ) = 0

=> ( a² - b² )³ + ( b² - c² )³ + ( c² - a² )³ = 3( a² - b² )( b² - c² )( c² - a² )

Similarly,

( a - b ) + ( b - c ) + ( c - a ) = 0

=> ( a - b )³ + ( b - c )³ + ( c - a )³ = 3( a - b )( b - c )( c - a )

Now, [ ( a² - b² )³ + ( b² - c² )³ + ( c² - a² )³ ] / [ ( a - b )³ + ( b - c )³ + ( c - a )³ ]

== 3( a² - b² )( b² - c² )( c² - a² ) / 3( a - b )( b - c )( c - a )

-Answered by Master Purushottam On 24 August 2019:12:09:53 AM(4857Average Rating Based on rating)

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