[tex]\it d) \ \ = 1\\ \\ \\ c)\ \ \Big(-\dfrac{5}{6}\Big)^2\cdot1\dfrac{1}{5}=\Big(\dfrac{5}{6}\Big)^2\cdot\dfrac{6}{5}=\dfrac{5}{6}\cdot\dfrac{\not5}{\not6}\cdot\dfrac{\not6}{\not5}=\dfrac{5}{6}\\ \\ \\ b)\ \ \Big(-\dfrac{1}{2}\Big)^3\cdot16=-\dfrac{1}{8}\cdot\dfrac{16}{1}=-2[/tex]
