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

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

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Find the cube root of $\frac{512}{3375}$.

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$(x)^{3} = 4913$. Find the value of $x$.

The smallest number by which $3600$ must be multiplied to make it a perfect cube is

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