Question

the formula of a3+b3+c3

Answer 1

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?

Answer 2

Concept:

Algebraic expressions are those expressins which has infinte roots i.e. true for all values.

(a+b)³=a³+b³+3ab(a+b)

(a+b)²=a²+b²+2ab

Given:

(a+b)=a³+b³+3ab(a+b)

Find:

a³ + b³ + c³=?

Solution:

a³ + b³ + c³ - 3abc = (a+b+c) (a² + b² + c² - ab - bc -

ac).

= (a+b+c) { (a²+b²+c² + 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 - 3a²b - 3ab² - 3abc - 3abc - 3b²c - 3bc² - 3a²c - 3abc - 3ac²

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

Therefore, a³ + b³ + c³ = (a+b+c)² - 3ab(a+b) - 3bc(b+c) - 3ca(c+a) - 9abc.

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