Let ℝ denote the set of all real numbers. Consider the polynomial...

Given $f(x) = \dfrac{d^{10}}{dx^{10}}\left((x^2 - 1)^{10}\right)$.

Using the binomial expansion, we have:
(x^2 - 1)^{10} = \sum_{k=0}^{10} (-1)^k \, ^{10}C_{k} x^{20-2k}

Differentiating $10$ times with respect to $x$, we get:
f(x) = \sum_{k=0}^{5} (-1)^k \, ^{10}C_{k} \dfrac{(20-2k)!}{(10-2k)!} x^{10-2k}

For the coefficient of $x^8$, we set $10 - 2k = 8 \Rightarrow k = 1$.
The coefficient is (-1)^1 \, ^{10}C_{1} \dfrac{18!}{8!} = -10 \left(\dfrac{18!}{8!}\right). Thus, statement (A) is true.

The highest power of $x$ in $f(x)$ corresponds to $k = 0$, which gives $x^{10}$ with a non-zero coefficient of $\dfrac{20!}{10!}$.
Therefore, the degree of the polynomial $f(x)$ is $10$. Thus, statement (C) is true.

For the constant term, we set $10 - 2k = 0 \Rightarrow k = 5$.
The constant term is (-1)^5 \, ^{10}C_{5} \dfrac{10!}{0!} = -\dfrac{10!}{5!5!} 10! = -\left(\dfrac{10!}{5!}\right)^2.
Thus, statement (D) is false.

To find $f(1)$ and $f(-1)$, we use the Leibniz rule for the $n$-th derivative of a product:
f(x) = \dfrac{d^{10}}{dx^{10}}\left((x-1)^{10}(x+1)^{10}\right) = \sum_{r=0}^{10} \, ^{10}C_{r} \dfrac{d^{10-r}}{dx^{10-r}}\left((x-1)^{10}\right) \dfrac{d^r}{dx^r}\left((x+1)^{10}\right)

Evaluating at $x = 1$, all terms in the sum are zero except when $r = 0$ (since $\dfrac{d^{10-r}}{dx^{10-r}}\left((x-1)^{10}\right)$ contains a factor of $(x-1)$ for $r > 0$).
f(1) = \, ^{10}C_{0} (10!) (1+1)^{10} = 10! \, 2^{10}

Evaluating at $x = -1$, all terms are zero except when $r = 10$.
f(-1) = \, ^{10}C_{10} (-1-1)^{10} (10!) = 10! \, 2^{10}

Adding these values gives:
$f(1) + f(-1) = 10! \, 2^{10} + 10! \, 2^{10} = 10! \, 2^{11}$
Thus, statement (B) is true.

Answer: The coefficient of $x^8$ in the polynomial $f(x)$ is $(-10)\left(\dfrac{18!}{8!}\right)$; The value of $f(1) + f(-1)$ is equal to $10!\, 2^{11}$; The degree of the polynomial $f(x)$ is $10$