Помогите с неравенствами 30.34 б,в

Решите задачу:

$\sqrt[6]{x^3-2x^2+1} \geq 0\; \; ,\; \; \; \; ODZ:\; \; x^3-2x^2+1 \geq 0\\\\x^3-2x^2+1=(x-1)(x^2-x-1)=(x-1)(x-\frac{1-\sqrt5}{2})(x-\frac{1+\sqrt5}{2}) \geq 0\\\\x\in [\frac{1-\sqrt5}{2},1]\cup [\frac{1+\sqrt5}{2},+\infty )\\\\x^3-2x^2+1 \geq 1^6\; \; (1^6=1)\\\\x^2(x-2) \geq 0\\\\Znaki:\; \; \; ---[0]--[2]+++\\\\x\in \{0\}\cup [\, 2,+\infty )$