Which one of the following is the common factor of $(47^{43} + 43^{43})$ and $(47^{47} + 43^{47})$?

### Concept & Formula This question tests the algebraic divisibility rule for the **sum of odd powers**. When raising two numbers to the same odd power and adding them, the result is always perfectly divisible by the sum of their bases. $$x^n + y^n \text{ is always divisible by } (x + y) \text{ when } n \text{ is an odd integer.}$$ ### Step-by-Step Solution * **Given:** We have two distinct mathematical expressions: $(47^{43} + 43^{43})$ and $(47^{47} + 43^{47})$. * **Calculation:** Let's evaluate the first expression: $(47^{43} + 43^{43})$. * Here, the exponent $n = 43$. Because 43 is an odd number, we can directly apply the binomial divisibility rule. The expression must be perfectly divisible by its base sum: $(47 + 43)$. * Now let's evaluate the second expression: $(47^{47} + 43^{47})$. * Here, the exponent $n = 47$. Because 47 is also an odd number, we apply the exact same rule. This expression is also perfectly divisible by its base sum: $(47 + 43)$. * Since $(47 + 43)$ is a confirmed factor for both independent expressions, it acts as their common factor. ### Exam Strategy & Shortcut Rote memorization of the three core binomial divisibility rules is mandatory for speed: 1. $x^n - y^n$ is divisible by $(x-y)$ for **all** $n$. 2. $x^n - y^n$ is divisible by $(x+y)$ for **even** $n$. 3. $x^n + y^n$ is divisible by $(x+y)$ for **odd** $n$. Spotting the "+" sign and the odd powers (43 and 47) instantly triggers rule 3, pointing straight to $(47 + 43)$ as the answer without writing down a single calculation. ### Common Pitfall Students often confuse the rules for sums and differences, mistakenly assuming that $x^n + y^n$ is universally divisible by $(x-y)$. This incorrect assumption leads many to confidently pick option (a) and lose easy marks. ### Final Answer Therefore, the correct answer is **$(47 + 43)$**.