32−15−27−2332^{-\frac15}-27^{-\frac23}32−51−27−32 A11/36B7/18C15/32D17/27Answer: B. 7/18 Read Explanation: 32−15−27−2332^{-\frac15}-27^{-\frac23}32−51−27−32 =(25)−15−(33)−23=(2^5)^{-\frac15}-(3^3)^{-\frac23}=(25)−51−(33)−32 =25×(−1/5)−33×(−2/3)=2−1−3−2=2^{5\times(-1/5)}-3^{3\times(-2/3)}=2^{-1}-3^{-2}=25×(−1/5)−33×(−2/3)=2−1−3−2 =12−19=\frac12-\frac19=21−91 =(9−2)18=7/18=\frac{(9-2)}{18}=7/18=18(9−2)=7/18 Read more in App