$(x^n - a^n)$ is divisible by $(x - a)$

## Concept & Formula According to the Remainder Theorem (and Factor Theorem), a polynomial $P(x)$ is divisible by $(x - a)$ if and only if $P(a) = 0$. ## Step-by-Step Solution * Let our polynomial be defined as $P(x) = x^n - a^n$. * We want to test divisibility by $(x - a)$. To do this, we find the root of the divisor by setting it to zero: $x - a = 0 \Rightarrow x = a$. * Substitute $x = a$ into our polynomial $P(x)$: $$ P(a) = (a)^n - a^n $$ $$ P(a) = a^n - a^n = 0 $$ * Because $P(a)$ evaluates to exactly 0, there is no remainder. * Notice that this subtraction equals 0 regardless of whether $n$ is odd, even, or prime. As long as $n$ is a natural number, the terms cancel out perfectly. ## Exam Strategy & Shortcut Substitute small integer values for $n$ to quickly verify the rule. * If $n = 1$: $(x^1 - a^1) = (x - a)$, which is obviously divisible by $(x - a)$. * If $n = 2$: $(x^2 - a^2) = (x - a)(x + a)$, which contains $(x - a)$ as a factor. * If $n = 3$: $(x^3 - a^3) = (x - a)(x^2 + ax + a^2)$, which contains $(x - a)$ as a factor. Since it works for both odd (1, 3) and even (2) numbers, it must be true for all values of $n$. ## Common Pitfall A frequent error is confusing the rules for $(x^n - a^n)$ with $(x^n + a^n)$. Remember that $(x^n + a^n)$ is only divisible by $(x + a)$ when $n$ is *odd*. However, with a negative sign in the middle, $(x^n - a^n)$ is unconditionally divisible by $(x - a)$. ## Final Answer **Therefore, the correct answer is for all values of $n$.**