Please note that a3=a3 , b3=b3 and c3=c3
Let f(a)= a3+b3+c3−3abc be a function in a.
Now, putting a = -(b+c), we get
f( -(b+c)) = b3+c3 - (b+c)3 + 3bc(b+c) = (b+c)3 - (b+c)3 = 0
Hence, by factor theorem, (a+b+c) is a factor.
Note that the expression a3+b3+c3−3abc is homogenous wrt a, b, and c hence no linear factors exist, as the degree of the expression is 3 and we have a factor of degree 1, the other factor must be of degree 2.
Hence we have the other factor = (a2 + b2 + c2) + k (ab + bc + ca) ; where k is any integer (since net coefficients are integers).
Now ((a2 + b2 + c2) + k (ab + bc + ca) ) (a+b+c) = a3+b3+c3−3abc.
The value of can be easily found out to be -1 (even by simply multiplying and comparing); hence the other factor, (a2 + b2 + c2 - ab - bc - ca) .
Thus a3+b3+c3−3abc = (a2 + b2 + c2 - ab - bc - ca) (a+b+c).
Note : The factor theorem usually helps in such questions only for finding homogenous factors or a single linear factor. It is partly guess work on what values should be put in f(a).
=[ a3 + b3 ]+c3 - 3abc
=[ (a+b)3 - 3ab(a+b) ] + c3 -3abc [ Put (a+b) = x ]
=x3 - 3abx + c3 - 3abc
=x3 + c3 - 3abx - 3abc
=(x + c)(x2 + c2 - cx) - 3ab(x + c)
=(x + c) [ (x2 + c2 - cx - 3ab) ] [ Replace x = a+b ]
=(a+b+c) [ (a+b) 2 + c2- c(a+b) - 3ab ]
=(a+b+c) [ a2+ b2+ 2ab + c2 - ca- cb - 3ab ]
=(a+b+c) [a2 + b2 + c2- ab - bc - ca ]