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Whole cube of (a+b+c) Prove |
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Answer» Answer: (a + b + c)³ = a³ + b³ + c³ + 3 (a +b) (b + c) (a+ c) Step-by-step explanation: 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. |
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