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Let $p(x)=a{x}^3+b{x}^2+cx+d.$ Then from the given information, $$\begin{array}{rl}a+b+c+d& =-7,\\ 8a+4b+2c+d& =-9,\\ 27a+9b+3c+d& =-15,\\ 64a+16b+4c+d& =-31.\end{array}$$Subtracting the first and second equations, second and third equations, and third and fourth equations, we get $$\begin{array}{rl}7a+3b+c& =-2,\\ 19a+5b+c& =-6,\\ 37a+7b+c& =-16.\end{array}$$Again subtracting the equations in pairs, we get $$\begin{array}{rl}12a+2b& =-4,\\ 18a+2b& =-10.\end{array}$$Subtracting once more, we get $6a=-6,$ so $a=-1.$ Back-substituting gives us $b=4,$ $c=-7,$ and $d=-3.$ Therefore, $$p(x)=\boxed{-x^3+4x^2-7x-3}.$$