To get faster at math, it is helpful to recognize the cube roots of perfect cubes from 1 to 10:
$\mathbf{1^3 = 1} \implies \sqrt[3]{1} = 1$
$\mathbf{2^3 = 8} \implies \sqrt[3]{8} = 2$
$\mathbf{3^3 = 27} \implies \sqrt[3]{27} = 3$
$\sqrt[3]{64} = \sqrt[3]{4 \times 4 \times 4} = \mathbf{4}$
$\sqrt[3]{125} = \sqrt[3]{5 \times 5 \times 5} = \mathbf{5}$
$\sqrt[3]{216} = \sqrt[3]{6 \times 6 \times 6} = \mathbf{6}$
$\sqrt[3]{343} = \sqrt[3]{7 \times 7 \times 7} = \mathbf{7}$
$\sqrt[3]{512} = \sqrt[3]{8 \times 8 \times 8} = \mathbf{8}$
$\sqrt[3]{729} = \sqrt[3]{9 \times 9 \times 9} = \mathbf{9}$
$\sqrt[3]{1000} = \sqrt[3]{10 \times 10 \times 10} = \mathbf{10}$
If you don’t have the cube root memorized, you can find it using the prime factorization method. Let’s look at an example.
Step 1: Perform Prime Factorization
Divide the number by the smallest possible prime numbers until you reach 1:
$216 \div 2 = 108$
$108 \div 2 = 54$
$54 \div 2 = 27$
$27 \div 3 = 9$
$9 \div 3 = 3$
$3 \div 3 = 1$
Step 2: Group the Factors into Triplets
Write the factors under the cube root symbol:
Step 3: Simplify
For every triplet of the same number, take one out of the radical:
To find the cube root of 2744, we begin by breaking it down into its prime factors:
Prime Factorization:
$2744 \div 2 = 1372$
$1372 \div 2 = 686$
$686 \div 2 = 343$
$343 \div 7 = 49$
$49 \div 7 = 7$
$7 \div 7 = 1$
Grouping: We group these factors into sets of three: $\sqrt[3]{(2 \times 2 \times 2) \times (7 \times 7 \times 7)}$
Solution: $\sqrt[3]{2744} = 2 \times 7 = \mathbf{14}$
For even larger numbers, the process remains the same.
Prime Factorization:
$35937 \div 3 = 11979$
$11979 \div 3 = 3993$
$3993 \div 3 = 1331$
$1331 \div 11 = 121$
$121 \div 11 = 11$
$11 \div 11 = 1$
Grouping: $\sqrt[3]{(3 \times 3 \times 3) \times (11 \times 11 \times 11)}$
Solution: $\sqrt[3]{35937} = 3 \times 11 = \mathbf{33}$