Local min

By rational root theorem , the polynomial can be factorized as $(8x + 1)(x-1)^2$ . Since 1 is the larger and the double root of this odd-degree polynomial with positive leading coefficient, then the local minimum occurs when $x=1$ . And so the answer is $8\cdot 1^3 - 15\cdot 1^2 + 6\cdot1 + 1 = \boxed0$ .

Let $f(x) = 8x^3 - 15x^2 + 6x+1$ . $f(x)$ has a local minimum, when $f'(x) = 0$ and $f''(x) > 0$ .

$\begin{aligned} f(x) & = 8x^3 - 15x^2 + 6x + 1 \\ f'(x) & = 24x^2 - 30x + 6 \end{aligned}$

Putting $f'(x) = 0$ , we have:

$\begin{aligned} 24x^2 - 30x + 6 & = 0 \\ 4x^2 - 5x + 1 & = 0 \\ (4x-1)(x-1) & = 0 \end{aligned}$

$\implies \begin{cases} f' \left( \frac 14 \right) = 0, & f'' \left( \frac 14 \right) = 48x - 30 \ \bigg|_{x = \frac 14} < 0 & \implies f \left(\frac 14 \right) \text{ is a maximum.} \\ f'(1) = 0, & f''(1) = 18 > 0 & \implies f(1) = \boxed 0 \text{ is a minimum.} \end{cases}$

I think the minimum of $\dfrac {8}{x^3}-\dfrac {15}{x^2}+\dfrac {6}{x}+1$ is asked. Differentiating this expression w. r. t. $x$ , equating to zero and solving we get $x=1,4$ . At $x=4$ , the expression is maximum, while at $x=1$ , it is minimum. The minimum value is $8-15+6+1=\boxed 0$ .