Lets look at the very similar equation:
$\displaystyle x^3 + 2.x^2 - 7·x - 14=0$
First thing to do is sketch the graph of:
$\displaystyle x^3 + 2.x^2 - 7·x - 14$
to see if we can find the approximate location of its zeros.
From the sketch in the attachment we see that this has zeros
near -2.7, -2, and +2.7. Trying these out we find that -2 is
an exact root of the cubic. So we may write:
$\displaystyle x^3 + 2.x^2 - 7·x - 14\ =\ (x+2)(A.x^2+B.x+C)$
Now you can use synthetic division to find the quadratic, which
in this case turns out to be:
$\displaystyle x^2-7$
which may be factorised either by inspection of by using the
quadratic formula to find its roots, and hence its linear factors.
So:
$\displaystyle x^3 + 2.x^2 - 7·x - 14\ =\ (x+2)(x+\sqrt 7)(x-\sqrt 7)$
So the solutions of the cubic are $\displaystyle -2,\ -\sqrt 7, +\sqrt 7$.
Your problem can be solved in a similar manner.
RonL