If $x = {(7 + 4 \sqrt{3})}^{2 n} = [x] + f$ , then $x (1 - f)$ is equal to ( [ ] denotes the integral part of $x$ ).......

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Solution

${(7 + 4 \sqrt{3})}^{2 n}$ = $^{2 n} C_{0}$ . $7^{2 n}$ + $^{2 n} C_{1}$ . $7^{2 n - 1}$ $4 \sqrt{3}$ .....................+ $^{2 n} C_{2 n - 1}$ . $7^{1}$ ${(4 \sqrt{3})}^{2 n - 1}$ + $^{2 n} C_{2 n}$ ${(4 \sqrt{3})}^{2 n}$
......eqn 1
let us assume $f =$ ${(7 - 4 \sqrt{3})}^{2 n}$ where $0 < f < 1$
${(7 - 4 \sqrt{3})}^{2 n}$ = $^{2 n} C_{0}$ . $7^{2 n}$ $-$ $^{2 n} C_{1}$ . $7^{2 n - 1}$ $4 \sqrt{3}$ ..................... $-$ $^{2 n} C_{2 n - 1}$ . $7^{1}$ ${(4 \sqrt{3})}^{2 n - 1}$ + $^{2 n} C_{2 n}$ ${(4 \sqrt{3})}^{2 n}$
....eqn 2
now adding eqn 1 and 2
${(7 + 4 \sqrt{3})}^{2 n}$ + ${(7 - 4 \sqrt{3})}^{2 n}$ $= 2$ (integer)
${(7 + 4 \sqrt{3})}^{2 n}$ $= I + F$ where $0 < F < 1$ ( is the fractional part of x)
and so $0 < f + F < 2$

and as ${(7 + 4 \sqrt{3})}^{2 n}$ + ${(7 - 4 \sqrt{3})}^{2 n}$ $= I + F + f = 2$ (integer)
so $F + f = 1$
$f = (1 - F) =$ ${(7 - 4 \sqrt{3})}^{2 n}$
$x (1 - f) =$ ${(7 + 4 \sqrt{3})}^{2 n}$ . ${(7 - 4 \sqrt{3})}^{2 n}$ $=$ ${(49 - 48)}^{2 n}$ $= 1$

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If x=(7+4√3)2n=[x]+f, then x(1−f) is equal to ( [ ] denotes the integral part of x).......

If $x = {(7 + 4 \sqrt{3})}^{2 n} = [x] + f$ , then $x (1 - f)$ is equal to ( [ ] denotes the integral part of $x$ ).......