Question

Which of the following numbers is a perfect cube?

Answer 1

Correct option is C. $$216$$

Firstly we have to find the factors for the above numbers

$$\sqrt[3]{141} =\sqrt[3]{(3\times47)}$$, Not a cube

$$\sqrt[3]{294} =\sqrt[3]{(2\times3\times7\times7)}$$ , not a cube

$$\sqrt[3]{216} =\sqrt[3]{(6\times6\times6)}$$, a cube

$$\sqrt[3]{496} =\sqrt[3]{(2\times2\times2\times2\times31)}$$, not a cube

From the above results, $$216$$ has the perfect cube factors.

$$\therefore$$ $$216$$ is the perfect cube.