pls explain how to do this activity in a simple way
Identity (a3 – b3) = (a – b) (a2 + ab + b2)
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.
- Take 27 cubes and place them to form a stack consisting of a 9 columns, each column consisting 3 cubes [fig. (i)].
- Remove one cube from this stack get a stack of 26 cubes (Arrangement I)
- 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.
- 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 - 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
Dear Student,
In this activity, you have take 27 cubes in such a way that can be placed by making 9 columns and each column consisting 3 cubes [fig. (i)].{here we have taken 3 cubes in each column because we need to calculate cube as given in the identity}
That would be equal to ---- Volume of stack in fig. 1(i) = a3
Now, if you remove one cube from this stack get a stack of 26 cubes (Arrangement I) { 27 -1=26}
Take out one cube that would be called 'b'.
Therefore, Volume of stack in fig. 1(ii) = b3
Hence, Volume of arrangement I = Volume of stack in fig. 1(i) – Volume of stack in fig. 1(ii) = a3 – b3
Now, arrangement II would be made with the help of 26 cubes. This arrangement consists of three stacks.
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).
Note
a = each column consisting 3 cubes
b = remove one cube from the stack
You can also perform this activity for any other values of a and b e.g. a = 6, b = 2.
ANOTHER WAY
1. Make a cuboid of dimensions (a–b) × a × a (b < a), using acrylic sheet and cellotape/adhesive as shown in Fig. 1.
2. Make another cuboid of dimensions (a–b) × a × b, using acrylic sheet and cellotape/adhesive as shown in Fig. 2.
3. Make one more cuboid of dimensions (a–b) × b × b as shown in Fig. 3.
4. Make a cube of dimensions b × b × b using acrylic sheet as shown in Fig. 4.
5. Arrange the cubes and cuboids made above in Steps (1), (2), (3) and (4) to obtain a solid as shown in Fig. 5, which is a cube of volume a3 cubic units.
Hope you got it.
Regards!
In this activity, you have take 27 cubes in such a way that can be placed by making 9 columns and each column consisting 3 cubes [fig. (i)].{here we have taken 3 cubes in each column because we need to calculate cube as given in the identity}
That would be equal to ---- Volume of stack in fig. 1(i) = a3
Now, if you remove one cube from this stack get a stack of 26 cubes (Arrangement I) { 27 -1=26}
Take out one cube that would be called 'b'.
Therefore, Volume of stack in fig. 1(ii) = b3
Hence, Volume of arrangement I = Volume of stack in fig. 1(i) – Volume of stack in fig. 1(ii) = a3 – b3
Now, arrangement II would be made with the help 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.
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).
Note
a = each column consisting 3 cubes
b = remove one cube from the stack
You can also perform this activity for any other values of a and b e.g. a = 6, b = 2.
ANOTHER WAY
1. Make a cuboid of dimensions (a–b) × a × a (b < a), using acrylic sheet and cellotape/adhesive as shown in Fig. 1.
2. Make another cuboid of dimensions (a–b) × a × b, using acrylic sheet and cellotape/adhesive as shown in Fig. 2.
3. Make one more cuboid of dimensions (a–b) × b × b as shown in Fig. 3.
4. Make a cube of dimensions b × b × b using acrylic sheet as shown in Fig. 4.
5. Arrange the cubes and cuboids made above in Steps (1), (2), (3) and (4) to obtain a solid as shown in Fig. 5, which is a cube of volume a3 cubic units.
Hope you got it.
Regards!