$2(a-1)^3 - 30(a-1)^2 + c(a-1) + 1 = 2a^3 - 30a^2 + ca = 2(a+1)^3 - 30(a+1)^2+c(a+1) - 1$ First equality gives $-6a^2+6a-2 + 60a-30 - c + 1 = 6a^2-66a+c+31=0$ Second equality: $6a^2+6a+2-60a-30+c-1 = 6a^2-54a+c-29=0$ Substracting first with second gives $12a=60$ or $a = 5$. So the consecutive $x$ values are $4,5,6$ This gives $c = -149$