Find $\sqrt[3]{216}$
Find $\sqrt[3]{512}$
Find $\sqrt[3]{729}$
Find $\sqrt[3]{1000}$
Find $\sqrt[3]{2744}$
Find $\sqrt[3]{4913}$
Find $\sqrt[3]{1728}$
Find $\sqrt[3]{3375}$
Find $\sqrt[3]{35937}$
Find $\sqrt[3]{1000000}$
Quick Tip: To find these cube roots manually, you can use prime factorization. For example, since $216 = 2^3 \times 3^3$, the cube root is $2 \times 3 = 6$
$\sqrt[3]{216} = 6$
Explanation: $216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3$. Grouping them into threes gives $2^3 \times 3^3$. Therefore, the cube root is $2 \times 3 = 6$.
$\sqrt[3]{512} = 8$
Explanation: Since $8 \times 8 = 64$ and $64 \times 8 = 512$, the cube root is exactly 8.
$\sqrt[3]{729} = 9$
Explanation: $729$ is $9^3$. You can also see this through factorization: $3 \times 3 \times 3 \times 3 \times 3 \times 3$ (or $3^6$). $\sqrt[3]{3^6} = 3^2 = 9$.
$\sqrt[3]{1000} = 10$
Explanation: This is a base-10 cube. Since there are three zeros, the cube root is 1 followed by one zero ($10 \times 10 \times 10$).
$\sqrt[3]{2744} = 14$
Explanation: Factorizing $2744$ gives $2^3 \times 7^3$. Multiplying the bases ($2 \times 7$) gives 14.
$\sqrt[3]{4913} = 17$
Explanation: $4913$ ends in 3, which only happens when the cube root ends in 7 ($7^3 = 343$). Testing $17 \times 17 \times 17$ confirms it is 4913.
$\sqrt[3]{1728} = 12$
Explanation: Factorization results in $2^6 \times 3^3$. Taking the cube root gives $2^2 \times 3$, which is $4 \times 3 = 12$.
$\sqrt[3]{3375} = 15$
Explanation: Any cube ending in 5 must have a root ending in 5. Since $10^3 = 1000$ and $20^3 = 8000$, the answer must be 15.
$\sqrt[3]{35937} = 33$
Explanation: The last digit is 7, so the root ends in 3. Looking at the “thousands” part (35), the nearest perfect cube below it is $27$ ($3^3$). This points to 33.
$\sqrt[3]{1000000} = 100$
Explanation: Similar to 1000, we count the zeros. There are six zeros, so the cube root will have $6 \div 3 = 2$ zeros. $100 \times 100 \times 100 = 1,000,000$.