Question

The value of $\left[\right(\sqrt{3}+\sqrt{2}{)}^6]$ is
( Here, $[.]$ represents the greatest integer function )

Answer 1

Let $(\sqrt{3}+\sqrt{2}{)}^6=I+f$
where $I$ is the integral part and $f$ is the fractional part.
As $0<\sqrt{3}-\sqrt{2}<1$
$\Rightarrow 0<(\sqrt{3}-\sqrt{2}{)}^6<1$
Assume $f^{\prime }=(\sqrt{3}-\sqrt{2}{)}^6$

We know that,
$(x+y{)}^6+(x-y{)}^6=2[{}^6{C}_0{x}^6{y}^0+{}^6{C}_2{x}^4{y}^2+{}^6{C}_4{x}^2{y}^4+{}^6{C}_6{x}^0{y}^6]$
Putting $x=\sqrt{3},y=\sqrt{2}$, we get
$(\sqrt{3}+\sqrt{2}{)}^6+(\sqrt{3}-\sqrt{2}{)}^6=2[{}^6{C}_0(\sqrt{3}{)}^6+{}^6{C}_2\left(\sqrt{3}{)}^4\right(\sqrt{2}{)}^2+{}^6{C}_4\left(\sqrt{3}{)}^2\right(\sqrt{2}{)}^4+{}^6{C}_6(\sqrt{2}{)}^6]=2[27+(15\times 9\times 2)+(15\times 3\times 4)+8]=2[35+270+180]=970\Rightarrow (\sqrt{3}+\sqrt{2}{)}^6+(\sqrt{3}-\sqrt{2}{)}^6=970$
So, $I+f+f^{\prime }=970\to \text{an integer}$
As $0<f<1$ and $0<f^{\prime }<1$
$\Rightarrow 0<f+f^{\prime }<2$
Also, $f+f^{\prime }$ should be an integer.
$\therefore f+f^{\prime }=1$
$I=970-1=969\therefore \left[\right(\sqrt{3}+\sqrt{2}{)}^6]=969$