I have been doing some maths and to continue on with the question I need to somehow solve:
$$ x^3-25x^2+200x-340 = 0 $$
How do I solve this, is there a formula that I can use or a method for solving it besides grouping?
(William Jagy) The three by three matrix is $$ \left( \begin{array}{ccc} a&b&c \\ d&e&f \\ g&h&i \end{array} \right) $$
Hint:
Set $z:=x+\dfrac{25}3$ to "depress" the polynomial to the form
$$z^3+pz+q=0.$$
Then set $z:=u-\dfrac p{3u}$ and plug in the equation:
$$0=u^3-pu+\frac{p^2}{3u}-\frac{p^3}{27u^3}+pu-\frac{p^2}{3u}+q=u^3-\frac{p^3}{27u^3}+q.$$
This is a quadratic equation in $u^3$,
$$(u^3)^2+q(u^3)-\frac{p^3}{27u^3}=0$$ which you solve in the usual way. In your case, the roots are real. Pick any of them and draw the value of $z$, then $x$.
After you have found this real root, divide the initial polynomial by $x-r$. This yields a quadratic polynomial, which will give you two conjugate complex roots, and you are done.
Note that the method is quite general, but it breaks when the equation in $u^3$ has complex roots. In this case, you need to resort to trigonometry as there is (provably) no algebraic formula.