Question

what is the solution of this question (a+b+c)³. ​

Answer 1

Answer:

(a+b+c)3=a3+b3+c3+(a+b+c)(ab+ac+bc).

Hope it helps....

Answer 2

Step-by-step explanation:

Let me help you with this formula in detail. (a + b + c) = a + b + c° + 3 (a +b) (b + c) (a+ c) Proof: %3D (a + b + c) = a + b³ + c³ + 3 (a +b) (b + c) (a+ c) It can be written as (a + b+ c)³ - a° - b3 - c° = 3 (a +b) (b + c) (a+ c) (1) %3D ......... Consider the L.H.S of equation (1), (a + b + c)³ - a - 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 %3D = 3 [ ab (a + b) + bc (b + c) + ac (a + c) + 2 abc ] = 3[ ab (a + b) + b?c + bc? + abc + a?c + ac? + abc ] = 3[ ab (a + b) + (abc + b?c) + (abc + a?c) + (bc? + ac?) ] = 3[ ab (a + b) + bc (a + b) + ac (a + b) + c? (a + b) ] = 3 [ (a + b) (ab + bc + ac + c?)] = 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.