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
$(a+2b)[b^2(a+b)^2+3a^2b(a+b)+a^4]$
$=(a+2b)[4a^2b^2+2ab^3+3a^3b+a^4+b^4]$
$=(4a^3b^2+2a^2b^3+3a^4b+a^5+ab^4)+(8a^2b^3+4ab^4+6a^3b^2+2ba^4+2b^5)$
$=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+2b^5$
 A: 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.
A: One of the intersting method for obtaining relations between elements of Fibonacci numbers is based on matrix method. I saw this post about a relation between Fibonacci numbers and now the peresent question. The strategy of answers of two questions are the same and is based on this fact that if $F_n=a$ and $F_n+1=b$ then $F_n+2=a+b$ and finish. Now I want to suggest the matrix method for this kind of Question.
Suppose that we want find a closed-form expression for $F_n^m$ where $F_n$ is the $n$th term of Fibonacci numbers and $m$ is non-negative integer number. The number $m$ is odd or even. Let $m=2k+1$ where $k$ is a natural number. We know the $Q$ matrix is in the following form 
$$
Q=
\left( \begin {array}{cc} 
0&1\\ 
1&1
\end {array} \right)
$$
And It is well-kown that the $n$th power of $Q$ matrix is as follows 
$$
Q^n=
\left( \begin {array}{cc} 
F_{{n-1}}&F_{{n}}\\ 
F_{{n}}&F_{{n+1}}
\end {array} \right)
$$
It is easy to see that 
$$
det(Q^n)={(-1)}^n \Longrightarrow F_{{n-1}}F_{{n+1}}-{F_{{n}}}^{2}={(-1)}^n
$$
Now, we want to find a closed form expression for $F_n^m=F_n^{2k+1}=F_n^{2k}\,F_n$. At first, we compute the following equation 
$$
{(F_{{n-1}}F_{{n+1}}-{F_{{n}}}^{2})}^k={(-1)}^{nk}
$$ 
We can rewrite the above eqation as shown
$$
F_n^{2k}=G(n)+{(-1)}^{nk} \Longrightarrow F_n^{2k+1}=F_n\,G(n)+F_n\,{(-1)}^{nk}
$$
Where $G(n)$ is a function based on the Fibonacci elements. For example, we want to find a closed form experesion for $F_n^5$. At first, we obtain the following equation 
$$
{(F_{{n-1}}F_{{n+1}}-{F_{{n}}}^{2})}^2={(-1)}^{2n}
$$
$$
{F_{{n-1}}}^{2}{F_{{n+1}}}^{2}+{F_{{n}}}^{4}-2\,F_{{n-1}}{F_{{n}}}^{2}
F_{{n+1}}=1
$$
$$
{F_{{n}}}^{4}=2\,F_{{n-1}}{F_{{n}}}^{2}
F_{{n+1}}-{F_{{n-1}}}^{2}{F_{{n+1}}}^{2}+1
$$
$$
{F_{{n}}}^{5}=2\,F_{{n-1}}{F_{{n}}}^{3}
F_{{n+1}}-{F_n}{F_{{n-1}}}^{2}{F_{{n+1}}}^{2}+F_n
$$
Now, by using the expression of ${F_{{n}}}^{5}$, we conclude that
$$
{F_{{n+1}}}^{5}=2\,F_{{n}}{F_{{n+1}}}^{3}
F_{{n+2}}-{F_{n+1}}{F_{{n}}}^{2}{F_{{n+2}}}^{2}+F_{n+1}
$$
 From the summation of the expressions  ${F_{{n}}}^{5}$ and ${F_{{n}}}^{5}$, we obtain our results. 
