Find permutations whose third power is known I have to find permutations $a$ such that $a^3=(1 \ 2)(3 \ 4)(5 \ 6)(7 \ 8 \ 9 \  10)$
 and I have to find at least 3 solutions.
So first I must find disjoint cycles.
                    Those are: 
 1 [2,2,2][4]      and   2 [4,2][4]  and 3[6][4] 
are there any more cycles?
and what is the order of numbers (1,2,3,...10) in those new cycles?
so is solution (1 2)(3 4)(5 6)(7 10 9 8) sufficient for 1 and (1 3 2 4)(5 6)(7 10 9 8) for 2?
 A: Yes for $p=(1 \ 2)(3 \ 4)(5 \ 6)(10 \ 9 \ 8 \ 7)$.
No for $p=(1 \ 3 \ 2 \ 4)(5 \ 6)(7 \ 10 \ 9 \ 8)$ ; it isn't a solution because $p^3$ would send $1$ onto $4$ instead of $2$.

Here is a way to find many solutions :
First, as you have well seen it, we must take for the last cycle $\color{red}{the \ reversed \  cycle \ (7 \ 10 \ 9 \ 8)}$ (because for a cycle $c$ on 4 elements, $c^4=id \ \iff \ (c^{-1})^3=c$, and there are no other solutions). 
A first global solution is :
$$ (1 \ 2)(3 \ 4)(5 \ 6)\color{red}{(7 \ 10 \ 9 \ 8)}$$
(You had recognized it). Please note that we use the fact that a transposition $t$ is such that $t^3=t$.
A family of 8 other solutions are found by considering an order-$6$ cycle on elements $1 \cdots 6$ :
$$ (\underline{1} \ 3 \ 6 \ \underline{2} \ 4 \ 5)\color{red}{(7 \ 10 \ 9 \ 8)}$$
$$ (\underline{1} \ 4 \ 6 \ \underline{2} \ 3 \ 5)\color{red}{(7 \ 10 \ 9 \ 8)}$$
$$ (\underline{1} \ 5 \ 3 \ \underline{2} \ 6 \ 4)\color{red}{(7 \ 10 \ 9 \ 8)}$$
$$ (\underline{1} \ 6 \ 3 \ \underline{2} \ 5 \ 4)\color{red}{(7 \ 10 \ 9 \ 8)}$$
$$ (\underline{1} \ 3 \ 5 \ \underline{2} \ 4 \ 6)\color{red}{(7 \ 10 \ 9 \ 8)}$$
$$ (\underline{1} \ 4 \ 5 \ \underline{2} \ 3 \ 6)\color{red}{(7 \ 10 \ 9 \ 8)}$$
$$ (\underline{1} \ 5 \ 5 \ \underline{2} \ 6 \ 6)\color{red}{(7 \ 10 \ 9 \ 8)}$$
$$ (\underline{1} \ 6 \ 5 \ \underline{2} \ 5 \ 6)\color{red}{(7 \ 10 \ 9 \ 8)}$$
with the following building recipe : 


*

*$1$ and $2$ must be separated by two elements,

*the same for $3$ and $4$,

*the same for $5$ and $6$.


(being understood that "separated" is by reference to a cyclic arrangement).
As the positions of $1$ and $2$ are "frozen", the choice is reduced to the $2\times2\times2 = 8$ solutions given upwards, to which we must add the exceptional first one.
Remarks :
a) about the necessity to gather $1,2 \cdots 6$ into a cycle :


*

*no solution can exist of the form $(1 a b c)(* * )$ for the reason seen upwards : we should have $p=(1 a b 2)(* *)$ but then $p^3=(1 2 a b)(* *)$ which is not what we desire.

*no solution can exist of the form $(a b c d)(1  2)$ for a similar reason.
b) As remarked by @Robert Shore, it is not evident that no other solution exists by grouping for example $7$ with other elements than $8,9,10$, even if our intimate conviction says that there none.
