Show that the equation represents an ellipsoid. Show that the equation $$\frac{4}{3}{x^2} + \frac{4}{3}{y^2} + \frac{4}{3}{z^2} + \frac{4}{3}{xy} + \frac{4}{3}{xz} + \frac{4}{3}{yz} = 1$$ represents an ellipsoid. Find the position and lengths of its principal half-axes.
Workings
$$(x+y)^2+(y+z)^2+(z+x)^2=\frac32\,.$$
Common factor $\left(\dfrac32\right)^{\frac12}$ not sure how to find length and position half-axes??

We have $$\frac32 = (x+y)^2 + (y+z)^2 + (x+z)^2 = 2x^2 + 2y^2 + 2z^2 + 2xy + 2yz + 2xy$$
so set $A =\begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2\end{bmatrix}$ and the equation can be rewritten as $$\left\langle A\begin{bmatrix} x \\ y \\ z\end{bmatrix},\begin{bmatrix} x \\ y \\ z\end{bmatrix}\right\rangle = \frac32$$
Now diagonalize $A$:
$$A= P^TDP = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\
 -\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{3}} \\
 0 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\\end{bmatrix}^T \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4\end{bmatrix}\begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\
 -\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{3}} \\
 0 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\\end{bmatrix}$$
so we have 
$$\frac32 = \left\langle A\begin{bmatrix} x \\ y \\ z\end{bmatrix},\begin{bmatrix} x \\ y \\ z\end{bmatrix}\right\rangle = \left\langle P^TDP\begin{bmatrix} x \\ y \\ z\end{bmatrix},\begin{bmatrix} x \\ y \\ z\end{bmatrix}\right\rangle = \left\langle DP\begin{bmatrix} x \\ y \\ z\end{bmatrix},P\begin{bmatrix} x \\ y \\ z\end{bmatrix}\right\rangle$$
Therefore we can define new coordinates $$\begin{bmatrix} x' \\ y' \\ z'\end{bmatrix} =  P\begin{bmatrix} x \\ y \\ z\end{bmatrix} = 
\begin{bmatrix}
 \frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}+\frac{z}{\sqrt{3}} \\
 \frac{z}{\sqrt{3}}-\frac{x}{\sqrt{2}} \\
 \frac{z}{\sqrt{3}}-\frac{y}{\sqrt{2}} \\
\end{bmatrix}$$
so we get $$\frac32 = \left\langle D\begin{bmatrix} x' \\ y' \\ z'\end{bmatrix},\begin{bmatrix} x' \\ y' \\ z'\end{bmatrix}\right\rangle = x'^2 + y'^2 + 4z'^2$$
or $$\left(\frac{x'}{\sqrt{\frac32}}\right)^2 + \left(\frac{y'}{\sqrt{\frac32}}\right)^2 + \left(\frac{z'}{\frac12\sqrt{\frac32}}\right)^2 = 1$$
