### $x + y = 1$, ${x^2} + {y^2} = 2$, ${x^{11}} + {y^{11}} = ?$

We have, $2xy = {(x + y)^2} - ({x^2} + {y^2}) = 1 - 2 = - 1$

Now, ${(x - y)^2} = {(x + y)^2} - 4xy = 1 + 2 = 3$

$x - y = \pm \sqrt 3$

Solving for x & y,

One of the two variables would be $\frac{{1 + \sqrt 3 }}{2}$ and the other would be $\frac{{1 - \sqrt 3 }}{2}$.

Now, ${x^{11}} + {y^{11}} = \frac{1}{{{2^{11}}}}[{(1 + \sqrt 3 )^{11}} + {(1 - \sqrt 3 )^{11}}]$

$= \frac{2}{{{2^{11}}}}({}^{11}{C_0} + {}^{11}{C_2} \times 3 + {}^{11}{C_4} \times 9 + {}^{11}{C_6} \times 27 + {}^{11}{C_8} \times 81 + {}^{11}{C_{10}} \times 243)$

$= \frac{1}{{{2^{10}}}}(1 + 165+2970+12474+13365+2673)$

$= \frac{{31648}}{{{2^{10}}}} = \frac{{989}}{{32}}$