Решить неравенство .

Решите задачу:

$t=x^3-x\\\\t+\sqrt{t+1} \geq 1\\\\\sqrt{t+1} \geq 1-t\\\\t+1 \geq 1-2t+t^2\\\\t^2-3t \leq 0\\\\t(t-3) \leq 0\; \; \; ++++[0]----[3]++++\\\\0 \leq t \leq 3\\\\ \left \{ {{x^3-x \geq 0} \atop {x^3-x \leq 3}} \right. \left \{ {{x(x-1)(x+1) \geq 0} \atop {x^3-x-3 \leq 0}} \right. \; \left \{ {{x\in [-1,0]U[1,\infty)} \atop {x^3-x-3\leq 0}} \right.$

$x^3-x-3=0\\\\x_1=\sqrt[3]{\frac{3}{2}-\sqrt{\frac{9}{4}-\frac{1}{27}}}+\sqrt[3]{\frac{3}{2}+\sqrt{\frac{9}{4}-\frac{1}{27}}}\approx 1,67\\\\x^3-x-3 \leq 0,x \leq x_1,x \leq 1,67\\\\Otvet:x\in [-1,0]U[1;1,67]$