An algebra problem by A Former Brilliant Member

Write $\begin{aligned} x & = \; \big(\sqrt{13} + \sqrt{11}\big)^6 \; = \; \sum_{j=0}^6 \binom{6}{j} 13^{\frac{6-j}{2}} 11^{\frac{j}{2}} \\ y & = \; \big(\sqrt{13} - \sqrt{11}\big)^6 \; = \; \sum_{j=0}^6 (-1)^j\binom{6}{j} 13^{\frac{6-j}{2}} 11^{\frac{j}{2}} \\ x + y & = \; \sum_{j=0}^6 \big(1 + (-1)^j\big)\binom{6}{j} 13^{\frac{6-j}{2}} 11^{\frac{j}{2}} \; = \; 2\sum_{j=0}^3 \binom{6}{2j} 13^{3-j} 11^j \end{aligned}$ so that $x +y$ is an integer while $0 < y < 1$ . Thus $\{x\} = 1 - y$ , and hence $x\big(1- \{x\}\big) \; = \; xy \; = \; \big[(\big(\sqrt{13}+\sqrt{11}\big)\big(\sqrt{13}-\sqrt{11}\big)\big]^6 \; = \; 2^6 \; = \; \boxed{64}$