Zero and n

Different path, same result:

Notice $a^2 = -4a -21$ and $a^3 = -4a^2 -21a$ , and $a^4 = -4a^3 - 21a^2$

Let $E = a^4 - a^3 + a^2$ , the expression in question.

Substitute for $a^4 = -4a^3 - 21a^2$ and get $E = -4a^3 - a^3 - 21a^2 + a^2 = -5a^3 - 20 a^2 = -5a(a^2+4a)$

Now substitute $-21 = a^2 + 4a$ to get $E = (-5a)(-21) = 105a$

Since $f(a) = 0$ , $\implies \red{a^2 + 4a + 21 = 0}$ . Then we have:

$\begin{aligned} a^4-a^3 + a^2 & = a^2(a^2-a+1) \\ & = a^2(\red{\cancel{a^2+4a+21}^0}-5a-20) \\ & = - 5a^2(a+4) \\ & = - 5a(a^2+4a) \\ & = -5a(\red{\cancel{a^2+4a+21}^0} - 21) \\ & = \boxed{105}a \end{aligned}$

Since $a$ is a root of the equation $x^2+4x+21=0$ , therefore

$a^2+4a+21=0\implies a^2-a+1=-5(a+1)\implies a^4-a^3+a^2=a^2(a^2-a+1)=-5a^2(a+1)=-5a(a^2+4a)=-5a(-21)=105a=na$ .

So $n=\boxed {105}$ .

As $a$ is it's root therefore, $f(a)=0$

$a^2+4a+21=0$

$\boxed{a^2=-(4a+21)}$

Multiplying both sides by $a$

$a^3=-(4a^2+21a)$

$a^3=-(-4(4a+21)+21a)=-(-16a+21a-84)$

$\boxed{a^3=84-5a}$

Multiplying both sides by $a$

$a^4=84a-5a^2=84a+5(4a+21)=84a+20a+105$

$\boxed{a^4=104a+105}$

Using these values of $a^2, a^3$ and $a^4$

$a^4-a^3+a^2=104a+105+5a-84-4a-21=\boxed{105a}$

Therefore the answer is $105$