Question

If ${(8+3\sqrt{7})}^n=\alpha +\beta$ where $n$ and $\alpha$ are positive integers and $\beta$ is a positive proper fraction,then

• A. $(1-\beta )(\alpha +\beta )=1$
• B. $(1+\beta )(\alpha +\beta )=1$
• C. $(1-\beta )(\alpha -\beta )=1$
• D. $(1+\beta )(\alpha -\beta )=1$

Answer 1

Answer: (A)

$(1-\beta )(\alpha +\beta )=1$

Given, $\alpha +\beta =(8+3\sqrt{7}{)}^n$

And

$1-\beta$ $=\frac{1}{\alpha +\beta }$ $=\frac{1}{(8+3\sqrt{7}{)}^n}$

Rationalizing, we get

$1-\beta$ $=\frac{1}{(8+3\sqrt{7}{)}^n}\frac{(8-3\sqrt{7}{)}^n}{(8-3\sqrt{7}{)}^n}$

$=\frac{(8-3\sqrt{7}{)}^n}{(64-63{)}^n}$ ........ $a^2-b^2=(a-b)(a+b)$

$=(8-3\sqrt{7}{)}^n$

Now
$(1-\beta )(\alpha +\beta )$

$=(8-3\sqrt{7}{)}^n(8+3\sqrt{7}{)}^n$

$=(64-63{)}^n$ ........ $a^2-b^2=(a-b)(a+b)$

$=1$

Option A is correct