Question

If b is the mean proportional between a and c, show that
abc(a + b + c)^3 = (ab + bc + ca)^3​

Answer 1

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,bc and  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