7. The real solution to the equation $3 x^{3}-[x]=3$ is
7. $x=\sqrt[3]{\frac{4}{3}} \quad$ (1) If $x \leqslant -1$, then $3 x^{3}-[x] \leqslant 3 x-x+13 x-x \geqslant 4$; (4) If $0 \leqslant x<1$, then $3 x^{3}$ $-[x]=3 x^{3}<3$; (5) If $1 \leqslant x<2$, then $3 x^{3}-[x]=3 x^{3}-1=3, x=\sqrt[3]{\frac{4}{3}}$. From this, we know the equation has only one solution $x=\sqrt[3]{\frac{4}{3}}$.