B and D.
B .
[tex]\frac{1}{x^{-1}\text{ }}=\text{ 1 }\times x^1=x[/tex]
D. Since all terms have the same base, we can add exponents.
[tex]x^{\frac{1}{3}}\times x^{\frac{1}{3}}\times x^{\frac{1}{3}}^{}=x^{(\frac{1}{3}+\frac{1}{3}+\frac{1}{3})}=x^1=x^{}[/tex]
