## CBSE Class 9 Maths Lab Manual – Algebraic Identity (a3 – b3) = (a – b) (a2 + ab + b2)

Objective
To verify the identity a3 – b3 = (a – b)(a2 + ab + b2) geometrically by using sets of unit cubes.

Prerequisite Knowledge
Volume of a cube = (Edge)3
Volume of a cuboid = l x b x h
a3 – b3 = (a – b)(a2 + ab + b2)

Materials Required
A set of 53 plastic or wooden cubes each of dimensions (1 x 1 x 1 unit)

Procedure
To verify a3 – b3 = (a – b)(a2 + ab + b2). Let a = 3 and b =1.

1. Take 27 cubes and place them to form a stack consisting of a 9 columns, each column consisting 3 cubes [fig. (i)].
2. Remove one cube from this stack get a stack of 26 cubes (Arrangement I)
3. Make arrangement II of 26 cubes. This arrangement consists of three stacks.
• The first stack consists of 18 cubes such as 9 columns of two cubes each. .
• The second stack consists of 6 cubes such as two rows of three cubes each.
• Third stack consist of 1 row of 2 cubes.

Observation
Since the two arrangements have equal number of cubes (each arrangement has 26 cubes), the total volume in both the arrangements must be equal.

1. Volume of arrangement I
Volume of stack in fig. 1(i) = a3
Volume of stack in fig. 1(ii) = b3
∴Volume of arrangement I = Volume of stack in fig. 1(i) – Volume of stack in fig. 1(ii) = a3 – b3
2. Volume of arrangement II
Volume of the stack in fig. 2 (i) = (a – b) a2
Volume of the stack in fig. 2(ii) = (a – b)ab
Volume of the stack in fig. 2 (iii) = (a – b)b2
Total volume of arrangement II = (a – b)a2 + (a – b)ab + (a – b)b2 = (a – b)(a2 + ab + b2).
Since number of cubes in arrangement I and II are equal.
∴a3 – b3 = (a – b)(a2 + ab + b2).

Result
The identity a3 – b3 = (a – b)(a2 + ab + b2) is verified geometrically by using cubes and cuboids.

Learning Outcome
In this way, students can learn the concept of verifying the identity geometrically by adding different cubes and cuboids.

Activity Time
Students can perform this activity for any other values of a and b e.g. a = 6, b = 2 and find volumes of different cubes and cuboids through this activity.

Viva Voce

Question 1.
What is the value of 33 – 23 ?
(3 – 2) (9 + 4 + 6) = 19.

Question 2.
What is the simplification of 123 – 23?
(12 – 2) (144 + 4 + 24) = 1720.

Question 3.
Factorize: 27x3 – 64y3.
(3x – 4y) (9x2 + 16y2 + 12xy).

Question 4.
Find 83 – 33
(5) (82 + 9 + 24) = 485.

Question 5.
Factorize: (2x + y)3 – (x + 2y)3.
(2x + y – x – 2y) [(2x + y)2 + (x + 2y)2 + (2x + y) (x + 2y)]
= (x – y) (7x2 + 7y2 + 13xy)

Question 6.
How many factors are possible of x3 – 27?
2 factors.

Question 7.
Write the real zero of 8x3 – 1.
$$\frac { 1 }{ 2 }$$

Question 8.
Is x – 2, a factor of x3 – 8?
Yes.

Question 9.
Flow many real zeroes are possible of x3 + 1 and x3 – 1 ?
Only one real zero is possible.

Multiple Choice Questions

Question 1.
Write the factors of a3 – b3:
(i) (a – b)(a2 + b2 + ab)
(ii) (a – b)(a2 + b2 – ab)
(iii) (a – b)(a + b)
(iv) none of these

Question 2.
Write the factors of 27 – x3:
(i) (3 – x)(9 + x2 + 3x)
(ii) (3 – x)(9 + x2 – 3x)
(iii) (x – 3)(9 + x2 + 3x)
(iv) none of these

Question 3.
Find the product of (3x – y) (9x2 + y2 + 3xy):
(i) 27x3 + y3
(ii) 27x – y3
(iii) y3 – 27x3
(iv) none of these

Question 4.
Write the degree of x3 – y3:
(i) 3
(ii) 2
(iii) 6
(iv) none of these

Question 5.
What is the coefficient of y in 343x3 – 27y3:
(i) -27
(ii) 343
(iii) 27
(iv) none of these

Question 6.
Write the coefficient of a3 in (125a3 – 8y):
(i) 125
(ii) -8
(iii) 8
(iv) none of these

Question 7.
If the area of rectangle is (125x3 -y3), find the dimensions of rectangle :
(i) (5x – y) and (5x + y)
(ii) (5x – y) and (25x2 + y2 – 5xy)
(iii) (5x – y) and (25x2 + y2 + 5xy)
(iv) none of these

Question 8.
Write the coefficient of a2 in (343a3 – 125b3):
(i) (7a – 5b)
(ii) (7a + 5b
(iii) (343 – 15b3)
(iv) none of these

Question 9.
64m3 – 343x3 is a
(i) binomial
(ii) monomial
(iii) trinomial
(iv) none of these

Question 10.
Evaluate: $$\frac { { 369 }^{ 3 }-{ 139 }^{ 3 } }{ { 369 }^{ 2 }+\left( { 369 }\times { 139 } \right) +{ 139 }^{ 2 } }$$
(i) 508
(ii) 230
(iii) 24508
(iv) none of these