Let $$f(x)$$ be a polynomial of degree 6 in $$x$$, in which the coefficient of $$x^6$$ is unity and it has extrema at $$x = -1$$ and $$x = 1$$. If $$\lim_{x \to 0} \frac{f(x)}{x^3} = 1$$, then $$5 \cdot f(2)$$ is equal to ______

Let $$f(x)$$ be a polynomial of degree 6 with leading coefficient 1. Since $$\lim_{x \to 0}\frac{f(x)}{x^3} = 1$$, the polynomial $$f(x)$$ must have no constant term, no $$x$$ term, and no $$x^2$$ term (otherwise the limit would be undefined or infinite), and the coefficient of $$x^3$$ must be 1.

So $$f(x) = x^6 + ax^5 + bx^4 + x^3$$ for some constants $$a$$ and $$b$$.

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

Computing the derivative: $$f'(x) = 6x^5 + 5ax^4 + 4bx^3 + 3x^2$$.

Setting $$f'(1) = 0$$: $$6 + 5a + 4b + 3 = 0$$, so $$5a + 4b = -9$$.

Setting $$f'(-1) = 0$$: $$-6 + 5a - 4b + 3 = 0$$, so $$5a - 4b = 3$$.

Adding these two equations: $$10a = -6$$, giving $$a = -\frac{3}{5}$$. Subtracting: $$8b = -12$$, giving $$b = -\frac{3}{2}$$.

Therefore, $$f(x) = x^6 - \frac{3}{5}x^5 - \frac{3}{2}x^4 + x^3$$.

Computing $$f(2) = 64 - \frac{3}{5}(32) - \frac{3}{2}(16) + 8 = 64 - \frac{96}{5} - 24 + 8 = 48 - \frac{96}{5} = \frac{240 - 96}{5} = \frac{144}{5}$$.

Therefore, $$5 \cdot f(2) = 5 \times \frac{144}{5} = 144$$.