Question

Here is the brainliest question :

Prove that (a+b+c)³ -a³ - b³ - c³ = 3(a+b)(b+c)(c+a)

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Need help for my elder sister.

Answer 1

To prove,

(a+b+c)³ - a³ - b³ - c³ = 3(a+b)(b+c)(c+a)

LHS = [ (a+b+c)³ - a³ ] - (b³ + c³)

= (a+b+c-a)[ (a+b+c)² + a² + a(a+b+c) ] - [ (b+c)(b²+c²+bc) ]

{ by using identity : a³ + b³ = (a+b)(a² + b² + -ab) and a³ - b³ = (a+b)(a² + b² + ab) }

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

= (b+c) [ b²+c²+3a²+ 3ab - b² - c² + 3bc ]

= (b+c) [ 3(a² + ab + ac + bc) ]

= 3 (b+c) [ a(a+b) + c(a+b) ]

= 3 (b+c) [ (a+c)(a+b) ]

= 3 (a+b)(b+c)(c+a) = RHS

Hence, proved.

Answer 2

hey friend here is your answer...
(a+b+C)3-a3-b3-c3=3(a+b)(b+C)(C+a)
(a+b)3+c3+3c(a+b)(a+b+C)-a3-b3-c3
a3+b3+3ab(a+b)+c3+3c(a+b)(a+b+C)-a3-b3-c3
a3+b3+c3+3ab(a+b)+3c(a+b)(a+b+C)-a3-b3-c3
3ab(a+b+3c(a+b)(a+b+C)3(a+b)[ab+C(a+b+C)]
3(a+b)[ab+ac+bc+c2]3(a+b)[a(b+C)+C(b+C)]
3(a+b)(b+C)(C+a)
see right side same right hence proved
hope it helps.......
please mark it as brainlist..........