Two Vectors and their Cross Product Let $v$ $=$ $\begin{pmatrix}0\\ c\\ 2c\end{pmatrix}$ and let $u$ $=$ $\begin{pmatrix}c^2-1\\ 2c\\ c-3\end{pmatrix}$. For what values of $c$ do the three vectors $v$, $u$ and the cross product of $v$ and $u$ form a basis of $R^3$.

I am mainly hoping to find a simpler way to do this.

I calculated the cross product to be:

$\begin{pmatrix}-3c^2-3c\\ 2c^3-2c\\ -c^3+c\end{pmatrix}$

Once I put all the vectors together, I tried to calculate the determinant of the matrix. But in the end, I got the following $6-degree$ characteristic polynomial:

$5c^6$ $-$ $c^4$ + $18c^3$ + $14c^2$

I know I am supposed to find the values of $c$ for which this polynomial does not equal $0$, but a polynomial of such large degree is very difficult to factorize. So i was just wondering if there is an easier way to do this problem. Is my current approach even correct?

Any help?
Excluding $c=0\implies\underline{u}=\underline{0}$:

As long as $\underline{u}$ and $\underline{v}$ are not parallel, then they together with $\underline{u}\times\underline{v}$ will form a basis for $\mathbb{R}^3$ since $\underline{u}\times\underline{v}$ is perpendicular to the plane containing $\underline{u}$ and $\underline{v}$.

Setting $\underline{u}=\lambda\underline{v}$ leads to the solution $c=-1$ only.

So $c$ can be any value apart from $0$ or $-1$.
Find which values of $c$ make $u$ and $v$ linearly independent (for instance you can row reduce a matrix with $u$ and $v$ as columns to see if it has full rank or not). Then $u\times v$ is necessarily orthogonal to both $u$ and $v$, and is therefore linearly independent with $u$ and $v$, so you get a basis for $\Bbb R^3$.