**Solution:**

We have given **a+b+c=0, a**^{3}+b^{3}+c^{3}=3, a^{5}+b^{5}+c^{5}=10, a^{4}+b^{4}+c^{4}=?

We know that (a+b+c)^{2 }= a^{2}+b^{2}+c^{2}+2(ab+bc+ac)

=> 0 = a^{2}+b^{2}+c^{2}+2(ab+bc+ac)

We have **(a+b+c)(a**^{2}+b^{2}+c^{2}) = a^{3}+b^{3}+c^{3}+a^{2}b+b^{2}a+b^{2}c+c^{2}b+c^{2}a+a^{2}c .........(1)

whereas,

(a+b+c)(ab+bc+ac) = a^{2}b+b^{2}a+b^{2}c+c^{2}b+c^{2}a+a^{2}c+3abc

**a**^{2}b+b^{2}a+b^{2}c+c^{2}b+c^{2}a+a^{2}c = -3abc .....(Since a+b+c = 0)................................(2)

Now from eqation (1) and (2)

0 = a^{3}+b^{3}+c^{3}-3abc

3abc = a^{3}+b^{3}+c^{3 }=3 (since a^{3}+b^{3}+c^{3}=3)

**abc=1 -------------------(3)**

**Now**

(a^{3}+b^{3}+c^{3})(**a**^{2}+b^{2}+c^{2})=**a**^{5}+b^{5}+c^{5}+a^{3}b^{2}+b^{3}c^{2}+c^{3}a^{2}+a^{2}b^{3}+b^{2}c^{3}+c^{2}a^{3} ---------------------(3)

a^{3}b^{2}+b^{3}c^{2}+c^{3}a^{2}+a^{2}b^{3}+b^{2}c^{3}+c^{2}a^{3 }= (a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})*(a+b+c)-(a^{2}b^{2}c+ab^{2}c^{2}+a^{2}bc^{2})

= 0-(a^{2}b^{2}c+ab^{2}c^{2}+a^{2}bc^{2})

= -3abc(ab+bc+ca)

= -3(ab+bc+ca) ......................(4)

Now form eqn. (3) and (4) we get

(a^{3}+b^{3}+c^{3})(**a**^{2}+b^{2}+c^{2})=**a**^{5}+b^{5}+c^{5}-3(ab+bc+ca)

3(**a**^{2}+b^{2}+c^{2})=10 - 3(ab+bc+ca)

3(-2(ab+bc+ac))=10-3(ab+bc+ca)

ab+bc+ac=-10/3

also

**3(a**^{2}+b^{2}+c^{2})=10+3/2(a^{2}+b^{2}+c^{2})

**a**^{2}+b^{2}+c^{2} = 20/3

(a^{2}+b^{2}+c^{2})^{2}=(20/3)^{2}

a^{4}+b^{4}+c^{4}+2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})=400/9

(ab+bc+ca)^{2}=a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+2(a^{2}bc+ab^{2}c+abc^{2} )

a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2 }=(-10/3)^{2 }- 2abc(a+b+c)=100/9

**a**^{4}+b^{4}+c^{4} = 400/9 - 2(100)/9=(400-200)/9=200/9

**a**^{4}+b^{4}+c^{4} = 200/9