jmc

Writing the equation as a quadratic in $a,$ we get $$a^2-(x^2+2x)a+(x^3-1)=a^2-(x^2+2x)a+(x-1)(x^2+x+1)=0.$$We can then factor this as $$(a-(x-1))(a-(x^2+x+1))=0.$$So, one root in $x$ is $x=a+1.$ We want the values of $a$ so that $$x^2+x+1-a=0$$has no real root. In other words, we want the discriminant to be negative. This gives us $1-4(1-a)<0,$ or $a<\frac{3}{4}.$

Thus, the solution is $a\in \boxed{\left(-\infty ,\frac{3}{4}\right)}.$