prove that (a+b+c)^3=a^3+b^3+c^3+3
(a+b)(b+c)(c+a)
Answers
Answer:
Let me help you with this formula in detail.
(a + b + c)³ = a³ + b³ + c³ + 3 (a +b) (b + c) (a+ c)
Proof:
(a + b + c)³ = a³ + b³ + c³ + 3 (a +b) (b + c) (a+ c)
It can be written as
(a + b + c)³ - a³ - b³ - c³ = 3 (a +b) (b + c) (a+ c) ......... (1)
Consider the L.H.S of equation (1),
(a + b + c)³ - a³ - b³ - c³
= a³ + b³ + c³ + 3 ab (a + b) + 3 bc (b + c) + 3 ac (a + c) +6 abc - a³ - b³ - c³
= 3 ab (a + b) + 3 bc (b + c) + 3 ac (a + c) +6 abc
= 3 [ ab (a + b) + bc (b + c) + ac (a + c) + 2 abc ]
= 3 [ ab (a + b) + b²c + bc² + abc + a²c + ac² + abc ]
= 3 [ ab (a + b) + (abc + b²c) + (abc + a²c) + (bc² + ac²) ]
= 3 [ ab (a + b) + bc (a + b) + ac (a + b) + c² (a + b) ]
= 3 [ (a + b) (ab + bc + ac + c²) ]
= 3 [ (a + b) { (c² + bc) + ( ab + ac) } ]
= 3 [ (a + b) { c ( b + c ) + a ( b + c ) } ]
= 3 (a + b) ( b + c) ( a + c )
which is equal to R.H.S of equation (1).
Thus proved.
Hope it will help you.
Thanks.
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