6. Given that for all $x \in[-1,1]$, we have $x^{3}-a x+1 \geqslant 0$.
Then the range of the real number $a$ is $\qquad$ .
6. $\left[0, \frac{3 \sqrt[3]{2}}{2}\right]$.

When $x=0$, the inequality obviously holds.
When $x \neq 0$, let $f(x)=x^{2}+\frac{1}{x}$.
If $x \in[-1,0)$, then the original inequality is equivalent to $f(x) \leqslant a$.
Note that, $f(x)$ is a decreasing function on the interval $[-1,0)$, and its maximum value is $f(-1)=0$.
Thus, $0 \leqslant a$.
If $x \in(0,1]$, then the original inequality is equivalent to
$$
\begin{array}{l}
f(x) \geqslant a . \\
\text { Also } x^{2}+\frac{1}{x}=x^{2}+\frac{1}{2 x}+\frac{1}{2 x} \\
\geqslant 3 \sqrt[3]{x^{2} \cdot \frac{1}{2 x} \cdot \frac{1}{2 x}}=\frac{3 \sqrt[3]{2}}{2},
\end{array}
$$

Therefore, the minimum value of $f(x)$ on the interval $(0,1]$ is
$$
\frac{3 \sqrt[3]{2}}{2} \geqslant a \text {. }
$$

In summary, the range of values for $a$ is $\left[0, \frac{3 \sqrt[3]{2}}{2}\right]$.