You will want to find the values of "c" and "a" so you must use the given values to find this.

Using the given values and some minor rearranging we can get following the 2 equations:

\(-3 = a+c\)

\(60 = 64a + c\)

We can now put these 2 equations into matrix form and rearrange to give the unknowns. See below.

\(\begin{pmatrix} 1 & 1\\ 64 & 1 \end{pmatrix} * \begin{pmatrix} a\\ c \end{pmatrix} = \begin{pmatrix} -3\\ 60 \end{pmatrix} \)

\(\therefore\begin{pmatrix} a\\ c \end{pmatrix} = {\begin{pmatrix} 1 & 1\\ 64 & 1 \end{pmatrix}}^{-1} * \begin{pmatrix} -3\\ 60 \end{pmatrix}\)

I used a calculator to perform this operation, I got:

\(a = 1, c = -4\)

to check that this works we can sub these values into one of the original equations, i'll use one for an example:

\(-5 = 1*({1}^{3})-2(1)-4\)

\(-5 = -5\)

This means that the values hold and the equation is:

\({x}^{3}-2x-4\)

Now to find the possible values of x you must allow the equation to equal zero and factor and use quadratic formula:

\(0 = {x}^{3} - 2x-4\)

factor out (x-2):

\((x-2)(x^2 + 2x + 2) = 0\)

from this we can gather taht a possible value for x is 2.

I wont proceed to do the calculations here, however using the quadratic formula, you can gather the remaining 2 possible answers.

the 3 possible values of x is:

\(x=2, x=-1+i, x=-1-i\)

hope this was helpful :)