Question

What are the factors of a^3 + b^3 + c^3 = 3abc ?
Plz guys help me its urgent

Answer 1

Let f(a)= a3+b3+c3−3.abca3+b3+c3−3.abc be a function in a.a.

Now, putting a=−(b+c)a=−(b+c), we get

f(−(b+c))=b3+c3−(b+c)3+3bc(b+c)=(b+c)3−(b+c)3=0f(−(b+c))=b3+c3−(b+c)3+3bc(b+c)=(b+c)3−(b+c)3=0

Hence, by factor theorem, (a+b+c)(a+b+c) is a factor.

Note that the expression a3+b3+c3−3abca3+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)=(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((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)(a2+b2+c2−ab−bc−ca) .

Thus (a3+b3+c3−3abc)=(a+b+c)(a2+b2+c2−ab−bc−ca).(a3+b3+c3−3abc)=(a+b+c)(a2+b2+c2−ab−bc−ca).

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).

Thanks

Answer 2

(a+b+c) (a sq. + b sq. + c sq. - ab - bc - ca)