ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{x^3+1}=a\ge0\Rightarrow x^3=a^2-1\)
\(B=\sqrt{a^2-1+2\left(a+1\right)}+\sqrt{a^2-1+2\left(1-a\right)}\)
\(=\sqrt{a^2+2a+1}+\sqrt{a^2-2a+1}\)
\(=\left|a+1\right|+\left|1-a\right|\ge\left|a+1+1-a\right|=2\)
\(B_{min}=2\) khi \(\left(a+1\right)\left(1-a\right)\ge0\Rightarrow0\le a\le1\Rightarrow-1\le x\le0\)