Rumination: Skip to the solution if you do not want to read what led me to the solution. I wonder if this problem would be rather easy if the condition given to us was simply this $(x-1)^3$ divides $p(x) .$ This would give us $p(1)=p'(1)=p''(1)=0$ and likewise for the other condition. But notice $ p(1)=-1$ but $ p'(1)=p''(1)=0. $ Ah, and there it is our solution.
Solution:
Since $p(x)+1$ is divisible by $(x-1)^3, $ therefore we must have
$$ p(x)+1 = q_{1}(x)(x-1)^3 $$
where $ q_{1}(x) $ is a polynomial of degree two. Notice that
$$ p(1) = -1, \ p'(1) = p''(1) = 0 \tag{A} $$
Similarly, since $p(x)-1$ is divisible by $(x+1)^3, $, therefore, we must have
$$ p(x)-1 = q_{2}(x)(x+1)^3 $$
where $ q_{2}(x) $ is a polynomial of degree two. Notice that
$$ p(-1) = 1, \ p'(-1) = p''(-1) = 0 \tag{B} $$
Since $p(x)$ is a polynomial of degree $5$, therefore, $p'(x)$ is a polynomial of degree $4$ whose zeros are $-1,1$ with multiplicity $2$ each.
$$ \implies p'(t) = A(t^2-1)^2 \implies \int_{-1}^{x} p'(t) dt = \int_{-1}^{x} A(t^2-1)^2 dt $$
$$ p(x)-p(-1) = p(x) -1 = A \left(\frac{x^5}{5}- \frac{2x^3}{3} + x + \frac{8}{15} \right) $$
Since $p(1)=-1,$ this gives us $A= \frac{-15}{8}$
$$ \implies p(x) = \frac{-3}{8} x^5 + \frac{5}{4}x^3 + \frac{-15}{8}x $$
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