Find the volume of the ellipsoid $11x^2+9y^2+15z^2-4xy+10yz-20xz=80$ 
Find the volume of the ellipsoid
  $$11x^2+9y^2+15z^2-4xy+10yz-20xz=80$$

Tried to make the expression simpler as
$$\frac{X^2}{A^2}+\frac{Y^2}{B^2}+\frac{Z^2}{C^2}=1$$
where $X, Y, Z$ are liner expressions of $x, y, z$. Then I would use the volume of a ball
$$\frac43\pi R^3$$
and write the volume of ellipsoid as
$$\frac43\pi ABC$$
But I'm stuck at the first step… any help?
 A: Because the ellipsoid has an equation involving only quadratic terms, it is centred at the origin and we can write the equation in matrix form as
$$\begin{bmatrix}x\\y\\z\end{bmatrix}^T\begin{bmatrix}11&-2&-10\\-2&9&5\\-10&5&15\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=80$$
The right-hand side is 80, so the eigenvalues $k_i$ of the coefficient matrix $A$ in the middle will be scaled by the same factor; the true eigenvalues $\lambda_i$ will be recovered when the $k_i$ are divided by 80. Now to find the eigenvalues of $A$:
$$\det(A-kI)=0$$
$$\begin{vmatrix}11-k&-2&-10\\-2&9-k&5\\-10&5&15-k\end{vmatrix}=0$$
$$(11-k)(9-k)(15-k)+100+100-100(9-k)-25(11-k)-4(15-k)=0$$
$$1485-399k+35k^2-k^3+200-900+100k-275+25k-60+4k=0$$
$$-k^3+35k^2-270k+450=0$$
We do not need to know the roots $k_1,k_2,k_3$ of this polynomial, only the product of the roots. By Viète's formulas we have
$$k_1k_2k_3=-\frac{450}{-1}=450$$
Dividing by $80^3$ we recover the product of the true eigenvalues:
$$\lambda_1\lambda_2\lambda_3=\frac{450}{80^3}=\frac9{10240}$$
Each eigenvalue is the square of the reciprocal of the length of one of the ellipsoid's semi-axes (see for example here, page 18). Therefore the product of the ellipsoid's semi-axis lengths is
$$abc=\lambda_1^{-\frac12}\lambda_2^{-\frac12}\lambda_3^{-\frac12}=(\lambda_1\lambda_2\lambda_3)^{-\frac12}=\sqrt{\frac{10240}9}=\frac{32\sqrt{10}}3$$
Finally, the ellipsoid's volume works out as
$$V=\frac43\pi abc=\frac43\pi\left(\frac{32\sqrt{10}}3\right)=\frac{128\sqrt{10}\pi}9=141.2919\dots$$
