Cubes and Cube Roots Class 8 PDF: Complete Guide & Practice

Understanding Cubes: Definition and Properties

The cube of a number is the product of that number multiplied by itself twice more. In simple terms, if $a$ is a number, then its cube is:

$$a^3 = a \times a \times a$$

For example, the cube of 3 is:

$$3^3 = 3 \times 3 \times 3 = 27$$

Key properties of cubes:

• Cubing a positive number gives a positive result.
• Cubing a negative number gives a negative result.
• The cube of zero is zero.

Memorizing cubes of numbers from 1 to 20 helps solve problems quickly. Here are a few:

Understanding these basics is essential for solving cube-related problems in Class 8 NCERT Maths.

Cube Roots: Meaning and Calculation Methods

The cube root of a number is the value that, when cubed, gives the original number. If $b$ is the cube root of $c$, then:

$$b = \sqrt[3]{c} \quad \text{such that} \quad b^3 = c$$

For example:

$$\sqrt[3]{27} = 3$$

because $3^3 = 27$.

Methods to find cube roots:

• Prime Factorization: Break the number into prime factors and group them in triples.
• Estimation: Use perfect cubes nearby to estimate.

Example: Find the cube root of 512 using prime factorization.

1. Prime factors of 512: $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$ (nine 2's) 2. Group in triples: $(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2)$ 3. Cube root is $2 \times 2 \times 2 = 8$

So, $\sqrt[3]{512} = 8$.

Formulas and Applications in Class 8 Mathematics

In the Class 8 NCERT chapter on cubes and cube roots, some important formulas and their applications include:

• Cube of a sum:

$$ (a + b)^3 = a^3 + b^3 + 3ab(a + b) $$

• Cube of a difference:

$$ (a - b)^3 = a^3 - b^3 - 3ab(a - b) $$

• Sum of cubes:

$$ a^3 + b^3 = (a + b)(a^2 - ab + b^2) $$

• Difference of cubes:

$$ a^3 - b^3 = (a - b)(a^2 + ab + b^2) $$

Application example:

Find the value of $27^3 - 26^3$ using the difference of cubes formula.

• Let $a = 27$, $b = 26$

$$ 27^3 - 26^3 = (27 - 26)(27^2 + 27 \times 26 + 26^2) $$

Calculate inside the bracket:

$$ 1 \times (729 + 702 + 676) = 1 \times 2107 = 2107 $$

So, $27^3 - 26^3 = 2107$.

These formulas simplify complex calculations and are crucial for CBSE Class 8 exams.

Solved Examples to Strengthen Your Concepts

Let's solve some typical problems from the cubes and cube roots chapter:

Example 1: Find the cube root of 1728.

Solution:

Prime factorize 1728:

$$1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$$

Group in triples:

$$ (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3) $$

Cube root is:

$$ 2 \times 2 \times 3 = 12 $$

So, $\sqrt[3]{1728} = 12$.

Example 2: Calculate $(4 + 3)^3$ using the cube of sum formula.

Solution:

Here, $a = 4$, $b = 3$

$$ (4 + 3)^3 = 4^3 + 3^3 + 3 \times 4 \times 3 \times (4 + 3) $$

Calculate each term:

• $4^3 = 64$
• $3^3 = 27$
• $3 \times 4 \times 3 \times 7 = 252$

Add all:

$$ 64 + 27 + 252 = 343 $$

Thus, $(4 + 3)^3 = 343$.

Practicing such examples from the NCERT textbook helps build confidence for exams.

Tips for Using the Cubes and Cube Roots Class 8 PDF Effectively

To make the most of your cubes and cube roots class 8 PDF, follow these tips:

• Start with definitions and formulas: Understand the basic concepts before attempting problems.
• Review solved examples: Study step-by-step solutions to grasp problem-solving methods.
• Practice all exercises: Complete NCERT textbook questions and additional worksheets.
• Use diagrams and tables: Visual aids help in memorizing cubes and cube roots.
• Revise regularly: Frequent revision improves retention and speeds up calculations.
• Attempt sample papers: Simulate exam conditions to test your knowledge.

By following these strategies, you can master the chapter and score well in your CBSE Class 8 Maths exams.