Functional value

$f(n)=a^n+b^n+c^n$

So,

$f(1)=a+b+c=1$ .......(1)

$f(2)=a^2+b^2+c^2=2$ ......(2)

$f(3)=a^3+b^3+c^3=3$ .......(3)

We know,

$ab+bc+ca=\dfrac{(a+b+c)^2-(a^2+b^2+c^2)}{2}$

i.e. $ab+bc+ca=\dfrac{(1)^2-2}{2}$

So,

$ab+bc+ca=\dfrac{-1}{2}$ .......(4)

Again we know,

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-(ab+bc+ca))$

$3-3abc=2+\dfrac{1}{2}$

$-3abc=\dfrac{-1}{2}$

$abc=\dfrac{1}{6}$

$36a^2b^2c^2=1$ ............(5)

From (2)×(3) we get,

$(a^2+b^2+c^2)(a^3+b^3+c^3)=6$

$a^5+b^5+c^5+(a^2b^3+a^2c^3+b^2a^3+b^2c^3+c^2a^3+c^2b^3)=6$

$a^5+b^5+c^5+(ba^2b^2+ca^2c^2+aa^2b^2+cb^2c^2+ac^2a^2+bb^2c^2)=6$

From (6)

$a^2b^2=1/36c^2$

$b^2c^2=1/36a^2$

$c^2a^2=1/36b^2$

so,

$a^5+b^5+c^5+(b/36c^2+c/36b^2+a/36c^2+c/36a^2+a/36b^2+b/36a^2)=6$

$a^5+b^5+c^5+{(a+b)/36c^2+(b+c)/36a^2+(c+a)/36b^2}=6$

$a^5+b^5+c^5+{(1-c)/36c^2+(1-a)/36a^2+(1-b)/36b^2}=6$

$a^5+b^5+c^5+{a^2b^2(1-c)+b^2c^2(1-a)+c^2a^2(1-b)}/36a^2b^2c^2=6$

$a^5+b^5+c^5+{a^2b^2+b^2c^2+c^2a^2-abc(ab+bc+ca)}=6$

$a^5+b^5+c^5+{a^2b^2+b^2c^2+c^2a^2+1/12}=6$ ........(6)

Now, squaring (4) we get,

$(ab+bc+ca)^2=1/4$

$a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)=1/4$

$a^2b^2+b^2c^2+c^2a^2+2*1/6*1=1/4$

$a^2b^2+b^2c^2+c^2a^2=-1/12$

From (6) we get,

$a^5+b^5+c^5+(-1/12+1/12)=6$

$a^5+b^5+c^5=6$

$f(5)=a^5+b^5+c^5=6$