SSC CGL 2022 PYQ — Question 73

The correct answer is Option 3 i.e. 2716.

 $p = 7+4\sqrt3$

So, after the rationalization

$\frac{1}{p} = 7-4\sqrt3$

p + 1/p = 14

p³ + 1/p³ = 14³ - 3 × 14 = 2702

The value of  ${{p^6+p^4+p^2+1} \over p^3}$ = $\frac{p^3(p^3+p+\frac{1}{p}+\frac{1}{p^3})}{p^3}$

= $(p^3+\frac{1}{p^3}+p+\frac{1}{p})$

= 2702 + 14 = 2716