krit.club logo

Cubes and Cube Roots - Finding Cube Roots by Prime Factorisation

Grade 8ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Definition of a Cube: A number xx is said to be the cube of a number nn if x=n×n×nx = n \times n \times n. Geometrically, this represents the volume of a solid cube with side length nn, where all three dimensions (length, breadth, and height) are equal.

Perfect Cubes: A natural number is called a perfect cube if it is the cube of some natural number. When expressed as a product of prime factors, every prime factor of a perfect cube appears in groups of three (triplets). Visually, think of a prime factor tree where every leaf node can be neatly bundled into sets of three identical numbers.

Properties of Cubes: The cubes of all even natural numbers are even (e.g., 23=82^3 = 8, 43=644^3 = 64), and the cubes of all odd natural numbers are odd (e.g., 33=273^3 = 27, 53=1255^3 = 125).

Cube Root Notation: The cube root of a number xx is the number nn such that n3=xn^3 = x. It is denoted by the radical symbol x3\sqrt[3]{x}. For example, since 43=644^3 = 64, we write 643=4\sqrt[3]{64} = 4.

Finding Cube Root by Prime Factorisation: This method involves three main steps: (1) Resolve the given number into its prime factors. (2) Group the identical factors into triplets. (3) Take one factor from each triplet and find their product to obtain the cube root. Visually, this is like decomposing a large block into smaller equal units.

Cube Roots of Negative Numbers: The cube root of a negative perfect cube is always negative. For any positive integer xx, x3=x3\sqrt[3]{-x} = -\sqrt[3]{x}. On a number line, this signifies moving to the left of zero by the same magnitude as the positive root.

Cube Roots of Product and Fractions: The cube root of a product of two numbers is equal to the product of their individual cube roots: a×b3=a3×b3\sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b}. Similarly, for fractions, ab3=a3b3\sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}, which allows for simplifying complex ratios step-by-step.

📐Formulae

n3=n×n×nn^3 = n \times n \times n

If x=n3x = n^3, then x3=n\sqrt[3]{x} = n

a×b3=a3×b3\sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b}

ab3=a3b3\sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}

x3=(x3)\sqrt[3]{-x} = -(\sqrt[3]{x})

(am)3=a3m(a^m)^3 = a^{3m}

💡Examples

Problem 1:

Find the cube root of 92619261 using the prime factorisation method.

Solution:

Step 1: Perform prime factorisation of 92619261. 9261=3×30879261 = 3 \times 3087 3087=3×10293087 = 3 \times 1029 1029=3×3431029 = 3 \times 343 343=7×49343 = 7 \times 49 49=7×749 = 7 \times 7 So, 9261=3×3×3×7×7×79261 = 3 \times 3 \times 3 \times 7 \times 7 \times 7

Step 2: Group the factors into triplets. 9261=(3×3×3)×(7×7×7)9261 = (3 \times 3 \times 3) \times (7 \times 7 \times 7)

Step 3: Take one factor from each triplet. 92613=3×7=21\sqrt[3]{9261} = 3 \times 7 = 21

Final Answer: 92613=21\sqrt[3]{9261} = 21.

Explanation:

We break down 92619261 into its smallest prime building blocks. Since 33 and 77 both appear exactly three times, we know 92619261 is a perfect cube. Multiplying the representative from each group gives the root.

Problem 2:

Determine the smallest number by which 675675 must be multiplied so that the product is a perfect cube.

Solution:

Step 1: Find the prime factorisation of 675675. 675=3×225675 = 3 \times 225 225=3×75225 = 3 \times 75 75=3×2575 = 3 \times 25 25=5×525 = 5 \times 5 So, 675=3×3×3×5×5675 = 3 \times 3 \times 3 \times 5 \times 5

Step 2: Identify the triplets. The prime factor 33 forms a complete triplet: (3×3×3)(3 \times 3 \times 3). The prime factor 55 only appears twice: (5×5)(5 \times 5).

Step 3: Find the missing factor. To make a triplet for 55, we need one more 55. Required number =5= 5.

Check: 675×5=3375675 \times 5 = 3375, and 33753=15\sqrt[3]{3375} = 15.

Explanation:

In prime factorisation, every factor must be in a group of three for the number to be a perfect cube. Here, the factor 55 is missing one member to complete its triplet.