Correct Answer: $\frac{\mathrm{k}^{64}-\frac{1}{\mathrm{k}^{64}}}{\mathrm{k}+\frac{1}{\mathrm{k}}}$

Solution : $
\left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}^8+\frac{1}{\mathrm{k}^8}\right)\left(\mathrm{k}^{16}+\frac{1}{\mathrm{k}^{16}}\right)\left(\mathrm{k}^{32}+\frac{1}{\mathrm{k}^{32}}\right)
$
Multiplying and dividing by $(\mathrm{k}+\frac{1}{\mathrm{k}})$
= $
\frac{(\mathrm{k}+\frac{1}{\mathrm{k}})\left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}^8+\frac{1}{\mathrm{k}^8}\right)\left(\mathrm{k}^{16}+\frac{1}{\mathrm{k}^{16}}\right)\left(\mathrm{k}^{32}+\frac{1}{\mathrm{k}^{32}}\right)}{(\mathrm{k}+\frac{1}{\mathrm{k}})}
$
= $
\frac{(\mathrm{k}^2-\frac{1}{\mathrm{k}^2})\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}^8+\frac{1}{\mathrm{k}^8}\right)\left(\mathrm{k}^{16}+\frac{1}{\mathrm{k}^{16}}\right)\left(\mathrm{k}^{32}+\frac{1}{\mathrm{k}^{32}}\right)}{(\mathrm{k}+\frac{1}{\mathrm{k}})}
$
= $
\frac{(\mathrm{k}^4-\frac{1}{\mathrm{k}^4})\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}^8+\frac{1}{\mathrm{k}^8}\right)\left(\mathrm{k}^{16}+\frac{1}{\mathrm{k}^{16}}\right)\left(\mathrm{k}^{32}+\frac{1}{\mathrm{k}^{32}}\right)}{(\mathrm{k}+\frac{1}{\mathrm{k}})}
$
= $
\frac{(\mathrm{k}^8-\frac{1}{\mathrm{k}^8})\left(\mathrm{k}^8+\frac{1}{\mathrm{k}^8}\right)\left(\mathrm{k}^{16}+\frac{1}{\mathrm{k}^{16}}\right)\left(\mathrm{k}^{32}+\frac{1}{\mathrm{k}^{32}}\right)}{(\mathrm{k}+\frac{1}{\mathrm{k}})}
$
= $
\frac{(\mathrm{k}^{16}-\frac{1}{\mathrm{k}^{16}})\left(\mathrm{k}^{16}+\frac{1}{\mathrm{k}^{16}}\right)\left(\mathrm{k}^{32}+\frac{1}{\mathrm{k}^{32}}\right)}{(\mathrm{k}+\frac{1}{\mathrm{k}})}
$
= $\frac{(\mathrm{k}^{32}-\frac{1}{\mathrm{k}^{32}})\left(\mathrm{k}^{32}+\frac{1}{\mathrm{k}^{32}}\right)}{(\mathrm{k}+\frac{1}{\mathrm{k}})}
$
= $\frac{(\mathrm{k}^{64}-\frac{1}{\mathrm{k}^{64}})}{(\mathrm{k}+\frac{1}{\mathrm{k}})} $
Hence, the correct answer is $\frac{(\mathrm{k}^{64}-\frac{1}{\mathrm{k}^{64}})}{(\mathrm{k}+\frac{1}{\mathrm{k}})}$.