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 :)

It is very easy to see what the minimal polynomial is for this matrix, even without knowing let alone factoring the characteristic polynomial. The argument even works for matrices of this type (called companion matrices) of arbitrary size $n$. Since $A\cdot e_i=e_{i+1}$ for $0<i<n$, where $e_i$ is the standard basis vector, it is clear that $(c_0I+c_1A+c_2A^2+\cdots+c_{n-1}A^{n-1})\cdot e_1=c_0e_1+c_1e_2+\cdots+c_{n-1}e_n$, which is obviously nonzero unless all the coefficients $c_i$ are zero. Therefore no nonzero polynomial of degree${}<n$ in$~A$ has $e_1$ in its kernel, and a fortiori no such polynomial can be the zero matrix. Therefore the minimal polynomial of$~A$ must have degree at least$~n$, so that it must coincide with the characteristic polynomial by Cayley-Hamilton.
You computed the characteristic polynomial in your example, but it would be a bit harder (though doable) for larger companion matrices. For the minimal polynomial it is easy by reasoning similarly to above. If the entries down the last column are $a_1,a_2,\ldots,a_n$, then $A\cdot e_n=a_1e_1+a_2e_2+\cdots+a_ne_n$. Now the minimal polynomial of$~A$ must be of the form $\mu=c_0+c_1X+\cdots+c_{n-1}X^{n-1}+X^n$, and computing as above one has $\mu[A]\cdot e_1=c_0e_1+c_1e_2+\cdots+c_{n-1}e_n+A^n\cdot e_1$, which must be zero; given that $A^n\cdot e_1=A\cdot e_n$ has the value just computed, it must be that $c_i+a_i=0$ for all $i$, so $c_i=-a_i$ fo all $i$ and hence $\mu=-a_0-a_1X+\cdots-a_{n-1}X^{n-1}+X^n$. You can conclude (using Cayley-Hamilton) that this is also the characteristic polynomial of$~A$. You can also prove this by direct calculation, either by induction on$~n$ and development of the determinant by the first row, or (without induction) by development of the determinant by the final column.
