minimal polynomial of a matrix with some unknown entries Question is to prove that  :
characteristic and minimal polynomial of $ \left( \begin{array}{cccc}
0 & 0 & c  \\
1 & 0 & b \\
0 & 1 & a \end{array} \right) $ is $x^3-ax^2-bx-c$.
what i have done so far is :
characteristic polynomial of a matrix $A$ is given by $\det(A-xI)$
in case of $A= \left( \begin{array}{cccc}
0 & 0 & c  \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)$ we have $\det(A-xI)=\det\left( \begin{array}{cccc}
-x & 0 & c  \\
1 & -x & b \\
0 & 1 & a-x \end{array} \right)=-(x^3-ax^2-bx-c)$
So, i have got the characteristic polynomial as $x^3-ax^2-bx-c$.
Now, the problem is how do i find minimal polynomial.
As $a,b,c$ are arbitrary, I can not factorize $x^3-ax^2-bx-c$ so as to see which factor gives me minimal polynomial.
I am confused.
please suggest me some hint. 
EDIT : This is just after Mr.Will Jagyy's hint :
I have $A= \left( \begin{array}{cccc}
0 & 0 & c  \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)$ then, $A^2= \left( \begin{array}{cccc}
0 & 0 & c  \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)\left( \begin{array}{cccc}
0 & 0 & c  \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)=\left( \begin{array}{cccc}
0 & c & ac  \\
0 & b & c+ab \\
1 & a & b+a^2 \end{array} \right)$
Now,
$A^2+rA+sI=\left( \begin{array}{cccc}
0 & c & ac  \\
0 & b & c+ab \\
1 & a & b+a^2 \end{array} \right)+r\left( \begin{array}{cccc}
0 & 0 & c  \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)+s\left( \begin{array}{cccc}
1 & 0 & 0  \\
0 & 1 & 0 \\
0 & 0 & 1 \end{array} \right)=\left( \begin{array}{cccc}
s & c & *  \\
r & b+s & * \\
1& * & * \end{array} \right)$
As element of $3^{rd}$ row $1^{st}$ column is $1$ in above matrix, this can never be $0$
i.e., $A^2+rA+sI$ can never be $0$.
Now, $A+rI=\left( \begin{array}{cccc}
0 & 0 & c  \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)+r\left( \begin{array}{cccc}
1 & 0 & 0  \\
0 & 1 & 0 \\
0 & 0 & 1 \end{array} \right)=\left( \begin{array}{cccc}
r & *& * \\
\color{magenta}{1} & * & * \\
* & \color{magenta}{1} & * \end{array} \right)\neq 0$ if $r\neq 0$
Thus, $A^2+rA+sI\neq 0$ and $A+rI\neq 0$ for any $r,s$.
Thus, minimal polynomial for $A$ can not be of order less than $3$.
Thus, minimal polynomial for $A$ has to be $x^3-ax^2-bx-c$.
I have written this just to make sure i have tried in correct way as i can not write this in a comment. 
I would be thankful if there is any other way to proceed further.. Thank you :)

Here's how I would argue the point:  clearly the minimal polynomial $m_A(x)$ of $A$,
where 
$A = \begin{bmatrix} 0 & 0 & c \\ 1 & 0 & b \\ 0 & 1 & a \end{bmatrix}, \tag{1}$
must satisfy $\deg m_A(x) \le 3$, since the characteristic poynomial $p_A(x)$ of $A$ satisfies $\deg p_A(x) = 3$.   Furthermore, we also have $m_A(x) \mid p_A(x)$.  Thus $p_A(x) = q(x)m_A(x)$ for some polynomial $q(x)$.  Now if $\deg m_A(x) = 3$, we must have $\deg q(x) = 0$ since $\deg p_A(x) = \deg q(x) + \deg m_A(x)$.  This, combined with the fact that $m_A(x), p_A(x)$ are both monic, leads to the conclusion that $q(x) = 1$ and hence $m_A(x) = p_A(x)$; we are done in this case.  Now suppose that $\deg m_A(x) = 1$; then $m_A(x)$ must be of the form $m_A(x) = x - d$ for some constant $d$.  Then since $m_A(A) = 0$, we obtain $A = dI$, where $I$ is the $3 \times 3$ identity matrix; then $A$ is diagonal; but this contradicts the fact that $A_{21} = A_{32} = 1$ for any choice of $a, b, c$; thus the possibility that $\deg m_A(x) = 1$ may be excluded.  The case $\deg m_A(x) = 2$ remains; here we have $m_A(x) = x^2 - dx - e$ for some $d, e$.  Then
$m_A(A) = 0$ implies $A^2 = dA + eI$.  We compute $A^2$:
$A^2 = \begin{bmatrix} 0 & 0 & c \\ 1 & 0 & b \\ 0 & 1 & a \end{bmatrix} \begin{bmatrix} 0 & 0 & c\\ 1 & 0 & b \\ 0 & 1 & a \end{bmatrix} =\begin{bmatrix}  0 & c & ac \\ 0 & b & ab + c \\ 1 & a & a^2 + b \end{bmatrix}, \tag{2}$
and observe that $A^2_{31} = 1$ for all $a, b, c$; on the other hand, inspection of (1) shows that, for any $d, e$,
$(dA + e)_{31} = 0; \tag{3}$
we thus conclude that $\deg m_A(x) = 2$ is impossible.  This leaves us with $\deg m_A(x) = 3$, and as we have seen, $m_A(x) = p_A(x)$ in this situation.  QED.
Additional Observations:  Some curiosities of the matrix $A$:  Suppose we write $A$ in the form
$A = A_1 + N, \tag{4}$
where
$A_1 = \begin{bmatrix} 0 & 0 & c \\ 0 & 0 & b \\ 0 & 0 & a \end{bmatrix}, \tag{5}$
and
$N = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}; \tag{6}$
then we have
$A_1^2 = \begin{bmatrix} 0 & 0 & ac \\ 0 & 0 & ab \\ 0 & 0 & a^2 \end{bmatrix}, \tag{7}$
and
$N^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix}; \tag{8}$
then
$A^2 = (A_1 + N)^2 = A_1^2 + A_1 N + N A_1 + N^2, \tag{9}$
with
$A_1 N = \begin{bmatrix} 0 & c & 0 \\ 0 & b & 0 \\ 0 & a & 0 \end{bmatrix} \tag{10}$
and
$N A_1 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & c \\ 0 & 0 & b \end{bmatrix} \tag{11}$
and
$N^3 = 0, \tag{12}$
i.e. $N$ is nilpotent of degree $3$; furthermore, (7) shows that the third column of $A_1$, $(c, b, a)^T$, is in fact an eigenvector of $A$ with eigenvalue $a$; thus
$A_1^m = \begin{bmatrix} 0 & 0 & a^{m - 1}c \\ 0 & 0 & a^{m - 1}b \\ 0 & 0 & a^m \end{bmatrix} \tag{13}$
readily follows; in contrast to (12), $A_1^m \ne 0$ as long as $a$ does not vanish.
Not sure where all this leads, but it seems that the peculiar form of $N$, $N^2$ etc.
has something to do with the above argument that $\deg m_A(x) \ne 2$. My guts tell me there is a generalization waiting to appear.  Like William Buther Yeats' "rough beast" (see his poem The Second Coming), something is "slouching towards Bethlehem to be born."
END:  Additional Observations:  Some curiosities of the matrix $A$.
Hope at least some of this spiel helps.  Cheerio,
and as always,
Fiat Lux!!!
