(10) Given positive real numbers $a$ and $b$ satisfying $a^{2}+b^{2}=1$, and $a^{3}+b^{3}+1=m(a+b+1)^{3}$, find the range of values for $m$.
(10) Let $a=\cos \theta, b=\sin \theta, 0<\theta<\frac{\pi}{2}$, then
$$
\begin{aligned}
m & =\frac{\cos ^{3} \theta+\sin ^{3} \theta+1}{(\cos \theta+\sin \theta+1)^{3}} \\
& =\frac{(\cos \theta+\sin \theta)\left(\cos ^{2} \theta-\cos \theta \sin \theta+\sin ^{2} \theta\right)+1}{(\cos \theta+\sin \theta+1)^{3}} .
\end{aligned}
$$

Let $x=\cos \theta+\sin \theta$, then $x=\sqrt{2} \sin \left(\theta+\frac{\pi}{4}\right) \in(1, \sqrt{2}]$, and
$$
\cos \theta \sin \theta=\frac{x^{2}-1}{2} \text {. }
$$

Thus,
$$
\begin{aligned}
m & =\frac{x\left(1-\frac{x^{2}-1}{2}\right)+1}{(x+1)^{3}}=\frac{2+3 x-x^{3}}{2(x+1)^{3}} \\
& =\frac{2+x-x^{2}}{2(x+1)^{2}}=\frac{2-x}{2(x+1)}=\frac{3}{2(x+1)}-\frac{1}{2} .
\end{aligned}
$$

Since the function $f(x)=\frac{3}{2(x+1)}-\frac{1}{2}$ is monotonically decreasing on $(1, \sqrt{2}]$, we have
$$
f(\sqrt{2}) \leqslant m<f(1) \text {. }
$$

Also, $f(1)=\frac{1}{4}, f(\sqrt{2})=\frac{3 \sqrt{2}-4}{2}$, so $m \in\left[\frac{3 \sqrt{2}-4}{2}, \frac{1}{4}\right)$.