Which of the following numbers is a perfect cube?
Correct option is B. $$1331$$
Firstly we have to find the factors for the above numbers
$$\sqrt[3]{1152} =\sqrt[3]{(2\times2\times2\times2\times2\times2\times2\times3\times3)} = 4\sqrt[3]{(2\times3\times3)}$$, not a cube
$$\sqrt[3]{1331} =\sqrt[3]{(11\times11\times11)}$$ , perfect cube
$$\sqrt[3]{2016} =\sqrt[3]{(2\times2\times2\times2\times2\times3\times3\times7)}=2\sqrt[3]{(2\times2\times3\times3\times7)}$$, not a cube
$$\sqrt[3]{739} =$$ no prime factors can be found, so not a cube
From the above results, $$1331$$ has the perfect cube factors.
$$\therefore$$ $$1331$$ is the perfect cube.