Proof for vectors involving cross and dot product

Yep, there is an easier way. Hint:

$$|a \times b| = |a||b|\sin(\theta)$$

and

$$a \cdot b = |a||b|\cos(\theta)$$

where $\theta$ is the angle between the vectors $a$ and $b$.

I prefer this purely vector method (using the Einstein summation convention) to the trig based method in the answer by Eli Rose. $$ \begin{align*} &\left| a\times b\right|^2+\left(a\cdot b\right)^2\\ =\;&\varepsilon_{ijk}a_jb_k\varepsilon_{ilm}a_lb_m+a_jb_ja_kb_k\\ =\;&a_jb_ka_lb_m(\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl})+a_jb_ja_kb_k\\ =\;&a_ja_jb_kb_k-a_jb_ka_kb_j+a_jb_ja_kb_k\\ =\;&a_ja_jb_kb_k=\left|a\right|^2\left|b\right|^2 \end{align*} $$ (Where $\varepsilon$ is the Levi-Civita symbol, $\delta$ is the Kronecker delta, and I've used the relation $\varepsilon_{ijk}\varepsilon_{ilm}=\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}$)