Solution:
We have given a+b+c=0, a3+b3+c3=3, a5+b5+c5=10, a4+b4+c4=?
We know that (a+b+c)2 = a2+b2+c2+2(ab+bc+ac)
=> 0 = a2+b2+c2+2(ab+bc+ac)
We have (a+b+c)(a2+b2+c2) = a3+b3+c3+a2b+b2a+b2c+c2b+c2a+a2c .........(1)
whereas,
(a+b+c)(ab+bc+ac) = a2b+b2a+b2c+c2b+c2a+a2c+3abc
a2b+b2a+b2c+c2b+c2a+a2c = -3abc .....(Since a+b+c = 0)................................(2)
Now from eqation (1) and (2)
0 = a3+b3+c3-3abc
3abc = a3+b3+c3 =3 (since a3+b3+c3=3)
abc=1 -------------------(3)
Now
(a3+b3+c3)(a2+b2+c2)=a5+b5+c5+a3b2+b3c2+c3a2+a2b3+b2c3+c2a3 ---------------------(3)
a3b2+b3c2+c3a2+a2b3+b2c3+c2a3 = (a2b2+b2c2+c2a2)*(a+b+c)-(a2b2c+ab2c2+a2bc2)
= 0-(a2b2c+ab2c2+a2bc2)
= -3abc(ab+bc+ca)
= -3(ab+bc+ca) ......................(4)
Now form eqn. (3) and (4) we get
(a3+b3+c3)(a2+b2+c2)=a5+b5+c5-3(ab+bc+ca)
3(a2+b2+c2)=10 - 3(ab+bc+ca)
3(-2(ab+bc+ac))=10-3(ab+bc+ca)
ab+bc+ac=-10/3
also
3(a2+b2+c2)=10+3/2(a2+b2+c2)
a2+b2+c2 = 20/3
(a2+b2+c2)2=(20/3)2
a4+b4+c4+2(a2b2+b2c2+c2a2)=400/9
(ab+bc+ca)2=a2b2+b2c2+c2a2+2(a2bc+ab2c+abc2 )
a2b2+b2c2+c2a2 =(-10/3)2 - 2abc(a+b+c)=100/9
a4+b4+c4 = 400/9 - 2(100)/9=(400-200)/9=200/9
a4+b4+c4 = 200/9