Question

Which of the following numbers are not perfect cubes? Given reasons in support of your answer:

• A. $648$
• B. $729$
• C. $8640$
• D. $8000$

Answer 1

Answer: (A), (C)

$648$
$8640$

$\sqrt[3]{3648}=\sqrt[3]{32^3\times 3^4}=6\sqrt[3]{33}$

$\sqrt[3]{3729}=\sqrt[3]{33^6}=3^2=9\sqrt[3]{38640}=\sqrt[3]{32^6\times 3^3\times 5}=12\sqrt[3]{35}\sqrt[3]{38000}\sqrt[3]{32^6\times 5^3}=20$

$648$ & $8640$ are not perfect cubes.