Question

a^3+b^3+c^3
please give me ans​

Answer 1

solution:-

There are mainy 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.

Hope that was basic enough.