10. Let the function $f(x)=4 x^{3}+b x+1(b \in \mathbf{R})$, for any $x \in[-1,1]$, it holds that $f(x) \geqslant 0$. Find the range of the real number $b$.
When $x=0$, the inequality obviously holds; when $x \in(0,1]$, $4 x^{3}+b x+1 \geqslant 0 \Rightarrow b \geqslant-4 x^{2}-\frac{1}{x}$. Let $g(x)=4 x^{2}+\frac{1}{x}$, $g^{\prime}(x)=8 x-\frac{1}{x^{2}}=\frac{8 x^{3}-1}{x^{2}}$, then $g(x)$ is monotonically decreasing on $\left(0, \frac{1}{2}\right)$ and monotonically increasing on $\left(\frac{1}{2}, 1\right)$, so $g(x)_{\min }=g\left(\frac{1}{2}\right)=3$, hence $b \geqslant-3$; when $x \in[-1,0)$, $4 x^{3}+b x+1 \geqslant 0 \Rightarrow b \leqslant-4 x^{2}-\frac{1}{x}$, since $g(x)$ is monotonically decreasing on $(-1,0)$, we have $g(x)_{\max }=g(-1)=3$, so $b \leqslant-3$. In summary, the range of real number $b$ is $\{-3\}$.