Question

what is the formula of a+b+c whole cube

Answer 1

Let me first remind what is (a+b+c)2(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)(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)(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=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)(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]a3+b3+c3=(a+b+c)3−[3a2(b+c)+3b2(a+c)+3c2(a+b)+6abc]

So, a3+b3+c3−3abca3+b3+c3−3abc

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

split the 9abc9abc among the three terms and now collect ab,bcab,bcand acac terms:

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

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

=(a+b+c)[a2+b2+c2+2ab+2bc+2ac−3ab−3bc−3ac]=(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]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+(a−c)2],a3+b3+c3−3abc=(a+b+c)12[(a−b)2+(b−c)2+(a−c)2],as (a−b)2=a2+b2−2ab(a−b)2=a2+b2−2ab etc.