Consider $x = (2+\sqrt[]{3})^6$, $x=[x]+t$, where $[x]$ is the integer part of $x$, and $t$ is the 'non integer' part of $x$. find $x(1-t)$ consider $x = (2+\sqrt[]{3})^6$, $x=[x]+t$, where $[x]$ is the integer part of $x$, and $t$ is the 'non integer' part of $x$. find the value of $x(1-t)$
 A: Note that $(2+\sqrt{3})^6+(2-\sqrt{3})^6$ is an integer, indeed an even integer. For imagine expanding each term, using the Binomial Theorem. The terms involving odd powers of $\sqrt{3}$ cancel.  
We have $2-\sqrt{3}=\frac{1}{2+\sqrt{3}}$. So $(2-\sqrt{3})^6=\frac{1}{x}$, and 
$$(2+\sqrt{3})^6+(2-\sqrt{3})^6=x+\frac{1}{x}.$$ 
Since $2-\sqrt{3}$ is between $0$ and $0.3$, the number $(2-\sqrt{3})^6$ is positive but close to $0$. So $x$ is close to but below the integer $x+\frac{1}{x}$, and therefore
$$\lfloor x\rfloor=x+\frac{1}{x}-1.$$  
Also, $t=x-(x+\frac{1}{x}-1)=1-\frac{1}{x}$, so $1-t=\frac{1}{x}$. It follows that $x(1-t)=(x)(1/x)=1$.
Remark: The result is obviously structural. In particular, the exponent $6$ is irrelevant. The same argument works with, for example, $(3+2\sqrt{2})^{99}$, and in many analogous situations. 
A crucial role was played by the conjugate $2-\sqrt{3}$ of the number $2+\sqrt{3}$. This sort of thing happens very frequently. 
The fact that  medium sized powers of $2+\sqrt{3}$ are almost integers (but just a little bit smaller than an integer) is at first a little startling. It may be enlightening to use the calculator to compute a few powers. For similar but not identical behaviour, compute some powers of $2+\sqrt{5}$.
