$2 \sqrt[3]{40} - 4 \sqrt[3]{320} + 3 \sqrt[3]{625} - 3 \sqrt[3]{5}$ is equal to

  • A
    $-2 \sqrt[3]{340}$
  • B
    $0$
  • C
    $\sqrt[3]{340}$
  • D
    $\sqrt[3]{660}$

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Similar Questions

The sum of the square of the first number and the cube of the second number is $568$. Also,the square of the second number is $15$ less than the square of $8$. What is the value of $\frac{3}{5}$ of the first number? (Assuming both numbers are positive)

Find the cube root of $0.000729$.

$9^{3} \times 81^{2} \div 27^{3} = (3)^{?}$

$\sqrt[3]{4663} + 349 = ? \div 21.003$

$(x)^{3} = 729$,find the value of $x$.

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