Let f(x) be a polynomial of degree 5 , and have extrema at x = 1...

Let the polynomial of degree $5$ be $f(x) = ax^5 + bx^4 + cx^3 + dx^2 + ex + k$.

Given $\displaystyle\lim_{x \to 0} \left(\dfrac{f(x)}{x^3}\right) = -5$, the terms of degree less than $3$ must be zero, and the coefficient of $x^3$ must be $-5$.

Thus, $k = 0$, $e = 0$, $d = 0$, and $c = -5$.

The polynomial becomes $f(x) = ax^5 + bx^4 - 5x^3$.

Differentiating with respect to $x$, we get:

$f'(x) = 5ax^4 + 4bx^3 - 15x^2$

Since $f(x)$ has extrema at $x = 1$ and $x = -1$, we have $f'(1) = 0$ and $f'(-1) = 0$.

$f'(1) = 5a + 4b - 15 = 0$

$f'(-1) = 5a - 4b - 15 = 0$

Adding both equations, we get $10a - 30 = 0 \Rightarrow a = 3$.

Subtracting the equations, we get $8b = 0 \Rightarrow b = 0$.

So, the polynomial is $f(x) = 3x^5 - 5x^3$.

Now, we find $f(2)$ and $f(-2)$:

$f(2) = 3(2)^5 - 5(2)^3 = 3(32) - 5(8) = 96 - 40 = 56$

$f(-2) = 3(-2)^5 - 5(-2)^3 = 3(-32) - 5(-8) = -96 + 40 = -56$

Therefore, $f(2) - f(-2) = 56 - (-56) = 112$.

Answer: $112$