Which of the following statement is correct?

I. If \(\mathrm{K}+\frac{1}{\mathrm{~K}}=12\), then \(\mathrm{K}^2+\frac{1}{\mathrm{~K}^2}=142\)

II. The value of \(\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}+\frac{1}{\mathrm{k}}\right) \) is

\(\mathrm{k}^{16}-\frac{1}{\mathrm{k}^{16}}\)

Given:

k + 1/k = 12

Formula used:

(x + 1/x)2 = (x2 + 1/x2) + 2

(a - b)(a + b) = a² - b²

Calculation:

Take first statement

k + 1/k = 12

using 

(x + 1/x)2 = (x2 + 1/x2) + 2

(k2 + 1/k2) = 144 - 2

(k2 + 1/k2) = =142

So, the first statement is true.

Take second statement

\(\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}+\frac{1}{\mathrm{k}}\right) \)

Using (a - b)(a + b) = a2 - b²

\(\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k^2}-\frac{1}{\mathrm{k^2}}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\)

\(\left(\mathrm{k^4}-\frac{1}{\mathrm{k^4}}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right) \)

\(\mathrm{k}^{8}-\frac{1}{\mathrm{k}^{8}}\)

So, the second statement is not true.

Therefore only statement I is true.

∴ Option 1 is the correct answer.

Mistake Points 

Please note that,

(K⁴)²  = K^{4 * 2} = K⁸

So statement 2 is not correct.