Question

prove that a cube + b cube + c cube - 3abc =1\2 (a+b+c) [(a-b) square +(b-c) square +(c-a) square ]

Answer 1

Hi ,

RHS=1/2(a+b+c)[(a-b)²+(b-c)²+(c-a)²]

=1/2(a+b+c)[a²-2ab+b²+b²-2bc+c²+c²-2ca+a²]

=1/2(a+b+c)[2a²+2b²+2c²-2ab-2bc-2ca]

=1/2(a+b+c)[2(a²+b²+c²-ab-bc-ca)]

= ( a + b + c )( a² + b² + c² - ab - bc - ca )

= a³ + b³ + c³ - 3abc

= LHS

Hence proved.

I hope this helps you.

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