Question

factorize a 3 +b 3 + c 3 +- 3abc

Answer 1

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).
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Answer 2

(a+b+c) (a^2+b^2+c^2-ab-bc-ca)