1. Let the cubic polynomial be $f(x)=ax^3+bx^2+cx+d$ Then, $f'(x)=3ax^2+2bx+c$ and $f''(x)=6ax+2b$

2. Use the condition: local minima of $f(x)$ at $x=1$

   For a local extremum at $x=1$, $f'(1)=0$ So, $3a+2b+c=0 \quad ...(1)$

   Since it is a local minimum, we should have $f''(1)>0$ which will be checked later.

3. Use the condition: local minima of $f'(x)$ at $x=-1$

   Since $f'(x)$ is a quadratic polynomial, its local minimum occurs at its vertex. Thus, $\frac{d}{dx}(f'(x))=f''(x)$ must vanish at $x=-1$: $f''(-1)=0$ Hence, $6a(-1)+2b=0$ $-6a+2b=0$ $b=3a \quad ...(2)$

   Also, for $f'(x)$ to have a local minimum, its quadratic coefficient must be positive: $3a>0 \Rightarrow a>0$

4. Use the given values

   From $f(1)=-10$, $a+b+c+d=-10 \quad ...(3)$

   From $f(-1)=6$, $-a+b-c+d=6 \quad ...(4)$

5. Substitute $b=3a$ into $(1)$

   $3a+2(3a)+c=0$ $3a+6a+c=0$ $c=-9a \quad ...(5)$

6. Now use equations $(3)$ and $(4)$

   From $(3)$: $a+3a-9a+d=-10$ $-5a+d=-10 \quad ...(6)$

   From $(4)$: $-a+3a-(-9a)+d=6$ $-a+3a+9a+d=6$ $11a+d=6 \quad ...(7)$

7. Solve for $a$ and $d$

   Subtract $(6)$ from $(7)$: $11a+d-(-5a+d)=6-(-10)$ $16a=16$ $a=1$

   Then, $b=3,\, c=-9$ and from $(6)$, $-5(1)+d=-10$ $d=-5$

   Therefore, $f(x)=x^3+3x^2-9x-5$

8. Check the minimum condition at $x=1$

   $f''(x)=6x+6$ $f''(1)=12>0$ So $x=1$ is indeed a local minimum.

9. Compute $f(3)$

   $f(3)=3^3+3(3^2)-9(3)-5$ $=27+27-27-5$ $=22$

10. Comparison with stored answer

Derived answer is $22$, which matches the stored correct answer.