Question

Prove that a^3 + b^3 + c^3 =3abc

Answer 1

Prove that a^3 + b^3 + c^3 =3abc

Your question is not fully correct. Given any three numbers a,b,cthe relation does not hold. However if a+b+c=0 or a=b=c you can prove the above relation is true. This is because (a+b+c)^3−3abc=(a+b+c)(a^2+b^2+c^2−ab−bc−ca).a^3+b^3+c^3−3abc=(a+b+c)(a^2+b^2+c^2−ab−bc−ca)

From this, a^3+b^3+c^3=3abc if either

a+b+c=0

or a=b=c,

since a^2+b^2+c^2−ab−bc−ca=[(a−b)^2+(b−c)^2+(c−a)^2]/2

We also see that the right hand side is a sum of squares, so it must always be non-negative, so a^3+b^3+c^3≥3abc

if and only if a+b+c≥0