# question from miscellaneous equations.

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if a+b+c=0, a3+b3+c3=3 and a5+b5+c5=10, then a4+b4+c4 is equal to

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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)= 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+c2a= (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+c2a=(-10/3)- 2abc(a+b+c)=100/9
a4+b4+c4 = 400/9 - 2(100)/9=(400-200)/9=200/9

a4+b4+c4 = 200/9