<s>[INST] Help to prove $F_n^5+F_{n+1}^5=F_{n+2}[(F_nF_{n+1}+F_{n-1}^2)^2+F_{n-1}^2F_nF_{n+1}]$ $0,1,1,2,3,5,8,...$ for $n=0,1,2,3,4...$ it is the n-th Fibonacci numbers.
$$F_n^5+F_{n+1}^5=F_{n+2}[(F_nF_{n+1}+F_{n-1}^2)^2+F_{n-1}^2F_nF_{n+1}]$$
How can we show that?
I try:
Expand and simplify to
$F_{n+2}[(F_nF_{n+1})^2+3F_{n-1}^2F_nF_{n+1}+F_{n-1}^4]=F_n^5+F_{n+1}^5$
Any hints would be helpful? Thank you!
I saw two to these identities from mathworld
$F_{n-1}F_{n+1}=F_n^2+(-1)^n$ and $F_n^4-F_{n-2}F_{n-1}F_{n+1}F_{n+2}=1$
May be these can help to simplify the above but I can't see it yet.
A hint from @Rohan
$(a+2b)[b^2(a+b)^2+3a^2b(a+b)+a^4]$
$=(a+2b)[4a^2b^2+2ab^3+3a^3b+a^4+b^4]$
$=(4a^3b^2+2a^2b^3+3a^4b+a^5+ab^4)+(8a^2b^3+4ab^4+6a^3b^2+2ba^4+2b^5)$
$=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+2b^5$
 [/INST] Hint: Consider $F_{n-1} = a, F_n =b$. Then, we get, $F_{n+1} = a+b$ and $F_{n+2} = a+2b$.   
Now then the LHS, becomes $[b^5 + (a+b)^5] = a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4 + 2b^5$. Try factoring this and the result follows. Hope it helps.
</s>