For all \(n \in N, a^{n}-b^{n}\) is always divisible by \((a\) and \(b\) are distinct rational numbers) (a) \(2 \mathrm{a}-\mathrm{b}\) (b) \(a+b\) (c) \(a-b\) (d) \(a-2 b\)

We need to find which of these options, (a), (b), (c), or (d), makes \(a^n - b^n\) always divisible by the option.

Check whether option (a) is true or false; that is, if \(a^n - b^n\) is always divisible by \(2a - b\). Let's test with the case when \(n = 1\). \(a^1 - b^1 = a - b\) \((a - b) \mod (2a - b)\) There is no guarantee for this equation to hold true. Counterexample: \(a=3, b=1\) \(a - b = 3 - 1 = 2\) \(2a - b = 6 - 1 = 5\) \(2 \not \equiv 0 \pmod{5}\) This counterexample shows that option (a) is not true.

Check whether option (b) is true or false; that is, if \(a^n - b^n\) is always divisible by \(a + b\). We will apply the identity \(x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + \cdots + y^{n-1})\). \(a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1})\) Now check if this expression is divisible by \(a + b\): \((a - b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1}) \mod (a + b)\) Since (a-b) and (a+b) are coprime, we just need to check if the second factor is divisible by (a+b). \((a^{n-1} + a^{n-2}b + \cdots + b^{n-1}) \mod (a + b) \equiv 0\) For odd values of \(n\), this equation holds true, so the expression is divisible by \(a + b\). For even values of \(n\), this property doesn't hold true in all cases. Thus, option (b) is also not true.

Check whether option (c) is true; that is, if \(a^n - b^n\) is always divisible by \(a - b\). We use the same identity as in step 3: \(a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1})\) Now check if this expression is divisible by \(a - b\): \((a - b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1}) \mod (a - b) \equiv 0\) As the factor \((a - b)\) is part of the expression itself, it is always divisible by \((a - b)\) for all \(n \in N\), and the given condition that \(a\) and \(b\) are distinct rational numbers. Therefore, option (c) is true.

We already found that option (c) is true, so there is no need to test for option (d). However, if you'd like to confirm it's false, the same counterexample used for option (a) will suffice, or you could use a similar approach as used in the previous steps.

Out of the four options provided, option (c) is the correct one: \(a^n - b^n\) is always divisible by \(a - b\) for all \(n \in N\) and given that \(a\) and \(b\) are distinct rational numbers.