Let x = (8 3 + 13) ^ 13 and y = (7 2 + 9) ^ 9 . If [t] denotes the...

We know the binomial expansion of ${\left(a + b\right)}^{n}$ is,

${\left(a + b\right)}^{n} = C0 n {a}^{n}{b}^{0} + C1 n {a}^{n - 1}{b}^{1} + C2 n {a}^{n - 2}{b}^{2} + . . . . . . + Cn n {a}^{0}{b}^{n}$

Using the above expansion,

$x = {\left(8\sqrt{3} + 13\right)}^{13} = C0 13 {\left(8\sqrt{3}\right)}^{13} + C1 13 {\left(8\sqrt{3}\right)}^{12}{\left(13\right)}^{1} + \ldots$

Let ${x}^{'} = {\left(8\sqrt{3} - 13\right)}^{13} = C0 13 {\left(8\sqrt{3}\right)}^{13} - C1 13 {\left(8\sqrt{3}\right)}^{12}{\left(13\right)}^{1} + \ldots$

So, $x - {x}^{'} = 2\left[C1 13 \cdot {\left(8\sqrt{3}\right)}^{12}{\left(13\right)}^{1} + C3 13 {\left(8\sqrt{3}\right)}^{10} \cdot {\left(13\right)}^{3} \ldots\right]$

Therefore, $x - {x}^{'}$ is an even integer, hence \left[x\right]$is even</p><p>Now,$y = {\left(7\sqrt{2} + 9\right)}^{9} = C0 9 {\left(7\sqrt{2}\right)}^{9} + C1 9 {\left(7\sqrt{2}\right)}^{8}{\left(9\right)}^{1} + C2 9 {\left(7\sqrt{2}\right)}^{7}{\left(9\right)}^{2} + . . . .$Now let,${y}^{'} = ( 7\sqrt{2} - 9{)}^{9} = C0 9 ( 7\sqrt{2}{)}^{9} - C1 9 ( 7\sqrt{2}{)}^{8} ( 9{)}^{1} + C2 9 {\left(7\sqrt{2}\right)}^{7}{\left(9\right)}^{2}$</p><p>So,$y - {y}^{'} = 2\left[C1 9 {\left(7\sqrt{2}\right)}^{8}{\left(9\right)}^{1} + C3 9 {\left(7\sqrt{2}\right)}^{6}{\left(9\right)}^{3} + \ldots\right]$</p><p>$y - {y}^{'} =$Even integer hence \left[y\right]$ is even