Răspuns :
[tex]\it a=\Big(\dfrac{1}{\dfrac{5}{3}}-\dfrac{1}{\dfrac{5}{6}}+\dfrac{1}{\dfrac{5}{2}}\Big)\cdot5\sqrt2=\dfrac{3-6+2}{5}\cdot5\sqrt2=-\dfrac{1}{5}\cdot5\sqrt2=-\sqrt2\\ \\ \\ a^3=(-\sqrt2)^3=(-\sqrt2)(-\sqrt2)(-\sqrt2)=-2\sqrt2[/tex]
[tex]\it a=\Big(\dfrac{1}{\dfrac{5}{3}}-\dfrac{1}{\dfrac{5}{6}}+\dfrac{1}{\dfrac{5}{2}}\Big)\cdot5\sqrt2=\dfrac{3-6+2}{5}\cdot5\sqrt2=-\dfrac{1}{5}\cdot5\sqrt2=-\sqrt2\\ \\ \\ a^3=(-\sqrt2)^3=(-\sqrt2)(-\sqrt2)(-\sqrt2)=-2\sqrt2[/tex]