Shorter way of evaluating the determinant of a $3\times 3$ Matrix 
Is there any short way of evaluating this?-I try to do it, but i end up making mistakes somewhere. The answer is supposed to be(I'm sorry I don't know how to resize the image):

But I never get this
 A: Hint: expand the determinant along the last row. You get the sum of two $2\times 2$ determinants with many common factors in the columns. You may perhaps factor out the factor $\frac15 r(v+5)\sin v$ from the second column first to make it even easier.
A: There are some common factors that can be pulled out of the determinant: in the first column there's $\frac 15$, in the middle there's $\frac 15r(v+5)\sin(v)$, in the last there's $r$, and in the last row there's $\frac {33}{20}$.  Then expand along the last row.
A: You can make life a bit easier by factoring out most of the nastiness:
$$\frac{33}{20}\left[\frac{v+5}{5}\right]^2 \left|\begin{matrix}
  \cos (u) \sin (v) & -r\sin(u) \sin( v) & r \cos(u)(\frac{\sin(v)}{v+5}+ \cos(v))\\
  \sin (u) \sin (v) & r\sin (v)\cos (v) & r \sin(u)(\frac{\sin(v)}{v+5}+ \cos(v)) \\
  \cos(v) & 0 & r\sin(v)
\end{matrix}\right|$$
A: You can do a Laplace expansion of a determinant along any row or column. If there are any zeroes in your matrix, choose the row or column with the most zeroes, so that you have less terms in the expansion.
For example, the last row of your matrix has a 0 in the second column, so I would choose to expand the determinant along that row (or the second column, your choice).
Using the last row, the Laplace expansion of the determinant is:
$$\begin{split}\frac{33}{20}\cos(v)\left|\begin{array}{cc}-\frac{1}{5}r(v+5)\sin(v)\sin(u) & r\cos(u)\left(\frac{\sin(v)}{5}+\left(\frac{1}{5}v + 1\right)\cos(v)\right)\\
\frac{1}{5}r(v+5)\sin(v)\cos(u) & r\sin(u)\left(\frac{\sin(v)}{5}+\left(\frac{1}{5}v + 1\right)\cos(v)\right)\end{array}\right| \\
- \frac{33}{20}r\sin(v)\left|\begin{array}{cc}\bigl(1 + \frac{1}{5}v\bigr)\cos(u)\sin(v) & -\frac{1}{5}r(v+5)\sin(v)\sin(u)\\
\frac{1}{5}(v+5)\sin(v)\sin(u) & \frac{1}{5}r(v+5)\sin(v)\cos(u)\end{array}\right|\end{split}$$
The determinants of the 2x2 matrices are straightforward. Factoring out common factors, we get:
$$\begin{split}\frac{33}{100}r^2(v+5)\sin(v)\left[-\cos(v)(\cos^2(u)+\sin^2(u))\left(\frac{\sin(v)}{5}+\left(\frac{1}{5}v+1\right)\cos(v)\right)\\ - \sin^2(v)\left(1+\frac{1}{5}v\right)\left(\cos^2(u)+\sin^2(u)\right)\right]\end{split}$$
Apply the Pythagorean identity $\sin^2(x)+\cos^2(x)=1$, expand the remaining binomial product in the first term within the square brackets, and pull out the common factor of $\frac{1}{5}$ to get:
$$\frac{33r^2}{500}(v+5)\sin(v)[-\sin(v)\cos(v)-(v+5)(\cos^2(v)+\sin^2(v))]$$
Applying the Pythagorean identity once more, we arrive at your expression, with the opposite sign. I believe your final answer does not have the correct sign, which my computer algebra system confirms.
