I have this so far:
From the problem, we know that $x^3$ is the leading term, and $3$ is the largest power. We also know that $f(x)$ has no constant because $f(0) = 0$. We can write our equation as $f(x) = a_3x^3 + a_2x^2 + a_1x$ with $a_3, a_2,$ and $a_1$ as coefficients of different powers of $x$. We can also factor $f(x) = a_3x^3 + a_2x^2 + a_1x$ into $f(x) = x(a_3x^2 + a_2x + a_1)$.
With the information $f(1) = -5$, we know that $a_3 + a_2 + a_1 = -5$ since $1^n = 1$.
Using the information $f(2) = 12$, we can plug in $2$ for $x$ in $f(x) = x(a_3x^2 + a_2x + a_1)$ to get $12 = 2(a_3(2^2) + a_2(2) + a_1)$. The equation simplifies into $12 = 2(4a_3 + 2a_2 + a_1)$, and by dividing both sides by $2$, we have $4a_3 + 2a_2 + a_1 = 12$.
Using the information $f(-1) = 15$, we can plug in $-1$ for $x$ in $f(x) = x(a_3x^2 + a_2x + a_1)$ to get $15 = -(a_3((-1)^2) + a_2(-1) + a_1)$. The equation simplifies into $15 = -(a_3 - a_2 + a_1)$, and by multiplying both sides by $-1$, we have $a_3 - a_2 + a_1 = -15$.
We have the three equations [#1] $a_3 + a_2 + a_1 = -5$, [#2] $4a_3 + 2a_2 + a_1 = 12$, and [#3] $a_3 - a_2 + a_1 = -15$. Using system of equations on the first and thrid equations, we have $(a_3 + a_2 + a_1) - (a_3 - a_2 + a_1) = -5 - (-15)$, which simplifies into $2a_2 = 10 \Rightarrow a_2 = 5$. Substituting $a_2 = 5$ in the first equation, we have [#4] $a_3 + a_1 = -10$, and substituting $a_2 = 5$ in the second equation, we have [#5] $4a_3 + a_1 = 2$. Using system of equations on the fourth and fifth equations, we have $(a_3 + a_1) - (4a_3 + a_1) = -10-2$, which simplifies into $-3a_3 = -12 \Rightarrow a_3 = 4$. Plugging in the values $a_2 = 5$ and $a_3 = 4$ into the first equation, we have $4 + 5 + a_1 = -5$, which simplifies to $a_1 = -14$.
Plugging the values $a_2 = 5$, $a_3 = 4$, and $a_1 = -14$ into $f(x) = x(a_3x^2 + a_2x + a_1)$, we have $f(x) = x(4x^2 + 5x -14)$.
