Mathematical physics problem. Eletric and Magnetic field. Velocity of propagation of the electromagnetic energy.

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If we define the velocity of propagation of the electromagnetic energy, for an arbitrary field in the vacuum, for $\vec{S} = U\vec{v}$ , where $U$ is the density of electromagnetic energy, then

$$\left(1-\frac{v^{2}}{c^{2}}\right)U^{2} = (U_{E} - U_{M})^{2} + \frac{\epsilon_{0}}{\mu_{0}}(\vec{E}.\vec{B})^{2}$$

I'm trying to use the equations: $$\vec{S}=\frac{1}{\mu_{0}}(\vec{E}\times\vec{B})$$ and $$(\vec{a}\times\vec{b})^{2} + (\vec{a}.\vec{b}) = a^{2}b^{2}$$

But I'm not getting that equality. Can someone help me?

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This is straightforward. Use $c^2=1/\sqrt{\epsilon_0\mu_0}$. \begin{align} \left(1-\frac{v^2}{c^2}\right)U^2&=U^2-\frac{v^2}{c^2}U^2=U^2-\frac{1}{c^2}\left\|\mathbf{S}\right\|^2\\ &=\left[\frac{1}{2}\left(\epsilon_0\left\|\mathbf{E}\right\|^2+\frac{1}{\mu_0}\left\|\mathbf{B}\right\|^2\right)\right]^2-\frac{1}{c^2}\left\|\frac{1}{\mu_0}\mathbf{E}\times\mathbf{B}\right\|^2\\ &=\frac{1}{4}\left(\epsilon_0^2\left\|\mathbf{E}\right\|^4+2\frac{\epsilon_0}{\mu_0}\left\|\mathbf{E}\right\|^2\left\|\mathbf{B}\right\|^2+\frac{1}{\mu_0^2}\left\|\mathbf{B}\right\|^4\right)-\frac{1}{c^2\mu_0^2}\left(\left\|\mathbf{E}\right\|^2\left\|\mathbf{B}\right\|^2-\left(\mathbf{E}\cdot\mathbf{B}\right)^2\right)\\ &=\frac{1}{4}\left(\epsilon_0^2\left\|\mathbf{E}\right\|^4+2\frac{\epsilon_0}{\mu_0}\left\|\mathbf{E}\right\|^2\left\|\mathbf{B}\right\|^2+\frac{1}{\mu_0^2}\left\|\mathbf{B}\right\|^4\right)-\frac{\epsilon_0}{\mu_0}\left(\left\|\mathbf{E}\right\|^2\left\|\mathbf{B}\right\|^2-\left(\mathbf{E}\cdot\mathbf{B}\right)^2\right)\\ &=\frac{1}{4}\left(\epsilon_0^2\left\|\mathbf{E}\right\|^4-2\frac{\epsilon_0}{\mu_0}\left\|\mathbf{E}\right\|^2\left\|\mathbf{B}\right\|^2+\frac{1}{\mu_0^2}\left\|\mathbf{B}\right\|^4\right)+\frac{\epsilon_0}{\mu_0}\left(\mathbf{E}\cdot\mathbf{B}\right)^2\\ &=\left[\frac{1}{2}\left(\epsilon_0\left\|\mathbf{E}\right\|^2-\frac{1}{\mu_0}\left\|\mathbf{B}\right\|^2\right)\right]^2+\frac{\epsilon_0}{\mu_0}\left(\mathbf{E}\cdot\mathbf{B}\right)^2. \end{align}