Question

a^3+b^3+c^3
slove it
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Answer 1

Answer:(a+b+c)^3

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

There are mainly two forms to write the formula of a³+b³+c³

First one is  

a³+b³+c³ = (a+b+c)[a²+b²+c²−ab−bc−ac]+3abc

Second one is

a³ + b³ + c³ = (a+b+c) 12[ (a−b)² + (b−c)² + (c−a)²] + 3abc

To derive these formulas, Follow these steps-

Let me first remind what is (a+b+c)²

(a+b+c)² = a²+b²+c² + 2(ab+bc+ac)(a+b+c)²= a² + b² + c² + 2(ab+bc+ac)

(a+b+c)³ = (a+b+c)² (a+b+c)

Thus after simplification we get,

(a+b+c)³ = a³ + b³ + c³ + a²(b+c) + b²(a+c) + c²(a+b) + 2(ab+bc+ac)(a+b+c)

=a³ + b³ + c³ + 3a²(b+c) + 3b²(a+c) + 3c²(a+b) + 6abc

So,

(a+b+c)³ = a³+b³+c³ + 3(a+b)(b+c)(a+c)

 

From the last but one step

a³+b³+c³ = (a+b+c)³ − [3a²(b+c)+3b²(a+c)+3c²(a+b)+6abc]

So,

a³ + b³ + c³ − 3abc = (a+b+c)3−[3a²(b+c)+3b²(a+c)+3c²(a+b)+9abc]

split the 9abc among the three terms and now collect ab,bcand ac terms:

=(a+b+c)³−[3ab(a+b+c)+3bc(a+b+c)+3ac(a+b+c)]

 

take (a+b+c) as common outside,

=(a+b+c) [a²+b²+c²+2ab+2bc+2ac−3ab−3bc−3ac]

 

Thus we get

a³ + b³ + c³ − 3abc = (a+b+c)[a²+b²+c²−ab−bc−ac]

which may further be rewritten as

a³ + b³ + c³ − 3abc = (a+b+c)12[(a−b)²+(b−c)²+(c−a)²]

 

as

(a−b)²=a²+b²−2ab etc.

Most important application of this identity comes when a+b+c=0.

Clearly the RHS=0, so

a³ + b³ + c³ = 3abc

Also let me add that the sign of the expression a³ + b³ + c³ − 3abc depends purely on the sign of a+b+c, as the term multiplied along is a perfect square and is always ≥0.