Six of a kind .

Finding patterns like this is known as the Prouhet-Tarry-Escott problem, see Wikipedia, also Chen, also Piezas.

For those who want the quick version, a particular example of Theorem 5 in the link cited by Zander states that if,

$$a^k+b^k+c^k = d^k+e^k+f^k$$

for $k=2,4$, then,

$$\small(x+a)^k+(x+b)^k+(x+c)^k+(x-a)^k+(x-b)^k+(x-c)^k = \\ \small (x+d)^k+(x+e)^k+(x+f)^k+(x-d)^k+(x-e)^k+(x-f)^k$$

for $k=1,2,3,4,5$. The example by the OP used,

$$5^k + 6^k + 11^k = 1^k + 9^k + 10^k$$

and $x=13$. However, to add a nice twist to this post, note the $6-10-8$ identity,

$$64\big(5^6 + 6^6 + 11^6 -(1^6 + 9^6 + 10^6)\big)\big(5^{10} + 6^{10} + 11^{10} -(1^{10} + 9^{10} + 10^{10})\big) =\\ 45\big(5^8 + 6^8 + 11^8 -(1^8 + 9^8 + 10^8)\big)^2$$

To know why, see this MO post.