Question

$\sqrt[3]{ x^{2}-4y }$

Answer 1

a)
para
x=13
y=11

$\sqrt[3]{x^2-4y} = \sqrt[3]{13^2-4(11)} = \\ \\ \sqrt[3]{169-44} = \sqrt[3]{125} = \\ \\ \sqrt[3]{5^3} =5$

b)
x=-9
y=27

$\sqrt[3]{(-9)^2-4(27)} = \sqrt[3]{81-98} = \sqrt[3]{-17}$