- -5

Please note that a3=a^{3} , b3=b^{3} and c3=c^{3}

Solution:

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).

- 3

^{3}+ b

^{3} + c

^{3}- 3abc

=[ a

^{3}+ b

^{3}]+c

^{3}- 3abc

=[ (a+b)

^{3}- 3ab(a+b) ] + c

^{3}-3abc [ Put (a+b) = x ]

=x

^{3} - 3abx + c

^{3 }- 3abc

=x

^{3}+ c

^{3 }- 3abx - 3abc

=(x + c)(x

^{2 }+ c

^{2 }- cx) - 3ab(x + c)

=(x + c) [ (x

^{2 }+ c

^{2 }- cx - 3ab) ] [ Replace x = a+b ]

=(a+b+c) [ (a+b)

^{2 }+ c

^{2}- c(a+b) - 3ab ]

=(a+b+c) [ a

^{2}+ b

^{2}+ 2ab + c

^{2 }- ca- cb - 3ab ]

=(a+b+c) [a

^{2 }+ b

^{2 }+ c

^{2}- ab - bc - ca ]

- 12