How to Find the Cube Roots of a Number?

The cube root of a real number x is such a number that when multiplied twice by itself gives us the number x. In other terms, the result generated by multiplying a number x three times is known as the cube of that number. The cube root of the integer x³ is x. The sign ∛x or (x)^1/3 represents a number's cube root. The prime factorization approach and the approximation method are both used to get the cube root of an integer. Let's take a closer look at both of them.

Prime Factorization Method

The prime factorization method is used to get the cube root of a perfect cube integer. It is also known as the long division method. A number having an integer cube root is called a perfect cube. It is a number that can be represented as the product of three integers that are all equal. 512, for instance, is a perfect cube since 8³ = 8 × 8 × 8 = 512. In this method, the given number is divided by its minimum value factors till the remainder becomes 1. The identical triplets are then grouped together and multiplied to get the value. If any factor cannot be grouped, it is taken inside the cube root which clearly indicates that the given number is not a perfect cube.

For example, 

We have to find the cube root of 512.

\begin{array} {|l}    \llap{2~~~~} 512 \\ \hline    \llap{2~~~~} 256  \\ \hline    \llap{2~~~~} 128  \\ \hline    \llap{2~~~~} 64  \\ \hline    \llap{2~~~~} 32   \\ \hline    \llap{2~~~~} 16    \\ \hline    \llap{2~~~~} 8    \\ \hline    \llap{2~~~~} 4    \\ \hline    \llap{2~~~~} 2    \\ \hline    \llap{} 1    \\ \end{array}

Write the prime factorization of 512.

512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

So, the cube root of the number becomes,

∛512 = ∛ (2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2)

∛512 = 2 x 2 x 2

∛512 = 8

Approximation Method

This method is used to determine the cube root of non-perfect cubes. It is also known as the estimation method. The important point to be noted while applying this method is that it doesn't give the exact value of the cube root of a number. Rather, it just provides a value close to the actual cube root. For example, the cube root of 4 is 1.587 but with this method, we get the cube root as 1.67. While applying this method, the prime factorization of the given integer is written first. All prime factors are added and divided by 3 to get an approximate value of the cube root.

For example,

We have to estimate the cube root of 3.

Express 3 in the form of its prime factors. So we get,

3 = 1 x 1 x 3

Cube root = (1 + 1 + 3)/3

= 5/3

= 1.67

This provides us the approximate value of cube root of 3, that is, 1.67. However its actual value is 1.442.

Sample Problems

Problem 1. Calculate the cube root of 64 by the prime factorisation method.

Solution:

We have to find the cube root of 64.

\begin{array} {|l}   \llap{2~~~~} 64 \\ \hline   \llap{2~~~~} 32  \\ \hline   \llap{2~~~~} 16  \\ \hline   \llap{2~~~~} 8    \\ \hline   \llap{2~~~~} 4    \\ \hline   \llap{2~~~~} 2    \\ \hline   \llap{~~~~} 1 \end{array}

Write the prime factorization of 64.

64 = 2 x 2 x 2 x 2 x 2 x 2

Taking the cube roots of both sides we get,

∛64 = ∛ (2 x 2 x 2 x 2 x 2 x 2)

= 2 x 2

= 4

Problem 2. Calculate the cube root of 343 by the prime factorisation method.

Solution:

We have to find the cube root of 343.

\begin{array} {|l}  \llap{7~~~~} 343 \\ \hline  \llap{7~~~~} 49 \\ \hline  \llap{7~~~~} 7  \\ \hline  \llap{~~~~}1 \end{array}

Write the prime factorization of 64.

343 = 7 x 7 x 7

Taking the cube roots of both sides we get,

∛343 = ∛ (7 x 7 x 7)

= 7

Problem 3. Calculate the cube root of 125 by the prime factorisation method.

Solution:

We have to find the cube root of 125.

\begin{array} {|l} \llap{5~~~~} 125 \\ \hline \llap{5~~~~} 25 \\ \hline \llap{5~~~~} 5  \\ \hline \llap{~~~~}1 \end{array}

Write the prime factorization of 125.

125 = 5 x 5 x 5

Taking the cube roots of both sides we get,

∛125 = ∛ (5 x 5 x 5)

= 5

Problem 4. Calculate the cube root of 729 by the prime factorisation method.

Solution:

We have to find the cube root of 729.

\begin{array} {|l} \llap{3~~~~} 729 \\ \hline \llap{3~~~~} 243 \\ \hline \llap{3~~~~} 81  \\ \hline \llap{3~~~~} 27  \\ \hline \llap{3~~~~} 9  \\ \hline \llap{3~~~~} 3  \\ \hline \llap{~~~~}1 \end{array}

Write the prime factorization of 729.

729 = 3 x 3 x 3 x 3 x 3 x 3

Taking the cube roots of both sides we get,

∛729 = ∛ (3 x 3 x 3 x 3 x 3 x 3)

= 3 x 3

= 9

Problem 5. Calculate the cube root of 216 by the prime factorisation method.

Solution:

We have to find the cube root of 216.

\begin{array} {|l} \llap{2~~~~} 216 \\ \hline \llap{2~~~~} 108 \\ \hline \llap{2~~~~} 54  \\ \hline \llap{3~~~~} 27  \\ \hline \llap{3~~~~} 9  \\ \hline \llap{3~~~~} 3  \\ \hline \llap{~~~~}1 \end{array}

Write the prime factorization of 216.

216 = 2 x 2 x 2 x 3 x 3 x 3

Taking the cube roots of both sides we get,

∛216 = ∛ (2 x 2 x 2 x 3 x 3 x 3)

= 2 x 3

= 6

Problem 6. Estimate the cube root of 2 by using the approximation method.

Solution:

We have to estimate the cube root of 2.

Express 2 in the form of its prime factors. So we get,

2 = 1 x 1 x 2

Cube root of 2 = (1 + 1 + 2)/3

= 4/3

= 1.33

Problem 7. Estimate the cube root of 5 by using the approximation method.

Solution:

We have to estimate the cube root of 5.

Express 5 in the form of its prime factors. So we get,

5 = 1 x 1 x 5

Cube root of 5 = (1 + 1 + 5)/3

= 7/3

= 2.33

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Article Tags :

• Mathematics
• School Learning
• Maths MAQ
• Maths-Formulas
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