Why a3 + b3 + c3 = (a+b+c)3 - 3ab(a+b) -3bc(b+c) -3ac(a+c) - 6abc?

[FlorenceGogan]

a³ - b³ - 3a²b + 3ab² = 2ac (a + b)³ = a³ + b³ + 3a²b + 3ab² (a - b)³

You can solve this question with the help of algebra-formulas.

a^3 + b^3 + c^3 - 3abc = (a+b+c) (a^2 + b^2 + c^2 - ab - bc -
ac).

= (a+b+c) { (a^2+b^2+c^2 + 2ab+2bc+2ac) - 3(ab+bc+ac) }

= (a+b+c) { (a+b+c)^2 - 3(ab+bc+ac) }

= (a+b+c)^2 - 3(ab+bc+ac)(a+b+c)

= (a+b+c)^2 - 3a^2b - 3ab^2 - 3abc - 3abc - 3b^2c - 3bc^2 - 3a^2c - 3abc - 3ac^2

= (a+b+c)^2 - 3ab(a+b) - 3bc(b+c) - 3ca(c+a) - 9abc

therefore,

a^3 + b^3 + c^3 = (a+b+c)^2 - 3ab(a+b) - 3bc(b+c) - 3ca(c+a)
- 6abc.

Can you tell about it, This answer is right or not and have a any other method for solving this question?

 

[stapel]

a³ - b³ - 3a²b + 3ab² = 2ac (a + b)³ = a³ + b³ + 3a²b + 3ab² (a - b)³

You can solve this question with the help of algebra-formulas.

Yes, we can. But what can you do? Until we've seen your work so far, I'm afraid there is little we can say, since we have no idea where you're getting stuck.

When you reply, please include the full and exact text of the exercise and the complete instructions, so we can see what you're actually being asked to do. (With three variables and only one equation, it is not possible to "solve" the equation.) Thank you!

 

[Deleted member 4993]

a^3 + b^3 + c^3 - 3abc = (a+b+c) (a^2 + b^2 + c^2 - ab - bc -
ac).

= (a+b+c) { (a^2+b^2+c^2 + 2ab+2bc+2ac) - 3(ab+bc+ac) }

= (a+b+c) { (a+b+c)^2 - 3(ab+bc+ac) }

= (a+b+c)^3 - 3(ab+bc+ac)(a+b+c)

= (a+b+c)^3 - 3a^2b - 3ab^2 - 3abc - 3abc - 3b^2c - 3bc^2 - 3a^2c - 3abc - 3ac^2

= (a+b+c)^3 - 3ab(a+b) - 3bc(b+c) - 3ca(c+a) - 9abc

therefore,

a^3 + b^3 + c^3 = (a+b+c)^3 - 3ab(a+b) - 3bc(b+c) - 3ca(c+a)
- 6abc.

Can you tell about it, This answer is right or not and have a any other method for solving this question?

correct ..... except the typos as indicated above

Another method (much longer) would be to use:

(a+b+c)³ = a³ + (b+c)³ + 3a(b+c)² + 3a²(b+c) .... and simplify ....