Solve for $x$ given the expression How do I get this expression in terms of $x$:
What I mean is having x = to something.
I just have no idea how to go about this problem now that I have it at this point.
$$x^3+c = 3x^2$$
Thanks for your help.
 A: Your equation is of the cubic type
$$x^3-3x^2+c=0$$
and was solved by Niccolò Tartaglia around 1530. The formulas are a little intricate.
A first step is to rewrite it in the form $x^3+px+q=0$, which is conveniently done in your case by dividing by $cx^3$ and letting $t=1/x$, $2s=1/c$ to get
$$t^3-6st+2s=0.$$
Then decompose $t=u+v$ and expand
$$t^3-6st+2s=u^3+3uv(u+v)+v^3-6s(u+v)+2s=0.$$
You get rid of the second and fourth terms by setting
$$uv=2s$$ and the equation simplifies to
$$u^3+v^3=-2s.$$
As $$u^3v^3=8s^3$$ you know the product and the sum of $u^3$ and $v^3$, which leads to a quadratic equation, and
$$u^3,v^3=-s\pm\sqrt{s^2-8s^3}.$$
Then taking the cubic roots give you $u,v$, then $t$ and $x$. To get all solutions (there are three of them), you need to resort to complex numbers when taking the cubic roots.
A: This equation is not solvable for $x$ in a nice form. In fact, if $c=1,$ then we get a cubic equation with rational coefficients (1) having no rational roots (2) and three real roots (3), and thus we're in the Casus irreducibilis situation for cubic equations, which means that none of the solutions can be expressed using a finite sequence of operations consisting of the four basic arithmetic operations along with the extraction of (positive integer) roots of real numbers. The solutions to $x^3 + 1 = 3x^2$ can be expressed using square and cube roots, but non-real complex numbers will be involved and the expressions cannot be simplified to remove the non-real numbers (unless you resort to a trigonometric or hyperbolic trig. presentation of the solution, or some such, rather than one that only makes use of arithmetic operations and extraction of roots).
(1) This is immediate. In fact, the coefficients are single digit integers.
(2) Use the precalculus rational root test to get this.
(3) By graphing $y=x^3 + 1$ and $y = 3x^2$ together (a very rough hand sketch is all that is needed), it is clear that there are at least two intersection points, and hence the cubic equation has at least two real solutions, and hence it must have three real solutions since any non-real solutions must occur in complex conjugate pairs (i.e. a cubic with real coefficients can have only one or three real solutions).
Of course, the situation you asked about, which involves an arbitrary (presumably real) parameter $c,$ is going to be much more complicated than the situation I discussed, where I simply put $c=1.$
