Prove that this equality is not achieved : $21+a^{2}+a^{bc}≠ a^{c+1}+a^{c+1}+(bc)^{a}$

Question

I need prove for all $a$ natural number such $a\geq 2$ and for all $b,c$ tow prime numbers then :

$$21+a^{2}+a^{bc}≠ a^{c+1}+a^{b+1}+(bc)^{a}$$

My attempts

I studied two cases :

1 • for $a\geq bc$ Then minimum values of $a$ is $4$ then For this reason, equality is incorrect So this situation is not possible.

2 • for $a\leq bc$

I don't know to complete

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I need some hints and thanks to all!