Question : If $\mathrm{p}=7+4 \sqrt{3}$, then what is the value of $\frac{\mathrm{p}^6+\mathrm{p}^4+\mathrm{p}^2+1}{\mathrm{p}^3}$?
Option 1: 2617
Option 2: 2167
Option 3: 2716
Option 4: 2176
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Correct Answer: 2716
Solution :
Given, $p=7+4\sqrt3$
⇒ $\frac1p=\frac{1}{7+4\sqrt3}$
⇒ $\frac1p=\frac{1}{7+4\sqrt3}\times\frac{7-4\sqrt3}{7-4\sqrt3}$
⇒ $\frac{1}{p}=\frac{7-4\sqrt3}{(7+4\sqrt3)(7-4\sqrt3)}$
We know $(a+b)(a-b)=a^2-b^2$
⇒ $\frac1p=\frac{7-4\sqrt3}{7^2-(4\sqrt3)^2}$
⇒ $\frac1p=\frac{7-4\sqrt3}{49-48}$
⇒ $\frac1p=7-4\sqrt3$
⇒ $p+\frac1p=7+4\sqrt3+7-4\sqrt3=14$
Now, consider $\frac{p^6+p^4+p^2+1}{p^3}$
$=\frac{p^6+1}{p^3}+\frac{p^4+p^2}{p^3}$
$=(p^3+\frac{1}{p^3})+(p+\frac{1}{p})$
We know, $(a+b)^3=a^3+b^3+3ab(a+b)$
So, $=(p^3+\frac{1}{p^3})+(p+\frac{1}{p})=(p+\frac1p)^3-3(p+\frac{1}{p})+(p+\frac1p)$
$=(p+\frac1p)^3-2(p+\frac{1}{p})$
$=14^3-2\times14$
$=2744-28$
$=2716$
Hence, the correct answer is 2716.
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