Can You Spot The Shortcut?

Notice that $13+21=34$ , $21+34=55$ , and so on...

All these numbers are part of the Fibonacci sequence , and in this particular scenario we can use the Gelin-Cesàro Identity , which states $\large{ F }_{ n }^{ 4 }-{ F }_{ n-2 }{ F }_{ n-1 }{ F }_{ n+1 }{ F }_{ n+2 }=1$ where ${F}_{n}$ is the $n$ th Fibonacci number and $n\ge2$ .

Thus ${34}^{4}-13\times21\times55\times89 =\large \color{#20A900}{\boxed{1}}$