Q:

Which monomial is a perfect cube?16x627x832x1264x6

Accepted Solution

A:
Answer:  The answer is (D) [tex]64x^6.[/tex]Step-by-step explanation:  We are to select the correct monomial from the given options which is a perfect cube.The first option is[tex]y=16x^6=2\times (8x^6)=2\times(2x^2)^3\\\\\Rightarrow \sqrt[3]{y}=2x^2\sqrt[3]{2}.[/tex]So, this is not a perfect cube.The second option is[tex]y=27x^8=x^2\times (27x^6)=x^2\times(3x^2)^3\\\\\Rightarrow \sqrt[3]{y}=3x^2\sqrt[3]{x^2}.[/tex]So, this is not a perfect cube.The third option is[tex]y=32x^{12}=4\times (8x^{12})=4\times(2x^4)^3\\\\\Rightarrow \sqrt[3]{y}=2x^4\sqrt[3]{4}.[/tex]So, this is not a perfect cube.The fourth option is[tex]y=64x^6=(4x^2)^3\\\\\Rightarrow \sqrt[3]{y}=4x^2.[/tex]So, this is a perfect cube.Thus, (D) [tex]64x^6[/tex] is the correct option.