NCERT Class 9 Maths Lab Manual – Verify the Algebraic Identity a³-b³ = (a-b) (a²+ab+b²)
To verify the algebraic identity a³-b³ = (a-b) (a²+ab+b²).
- Acrylic sheet
- Geometry box
- Adhesive/Adhesive tape
- Concept cuboid and its volume.
- Concept of cube and its volume.
- For concept of cuboid and its volume refer to Activity 7.
- For concept of cube and its volume refer to Activity 7.
- Using acrylic sheet, make a cuboid of dimensions (a – b) x a x a, where b < a. (see Fig. 10.1)
- Using acrylic sheet, make another cuboid of dimensions (a-b) x a x b, where b < a. (see Fig. 10.2).
- Now, make one more cuboid of dimensions (a-b) x b x b. (see Fig. 10.3)
- Now, make a cube of dimensions b x b x b. (see Fig. 10.4)
- Arrange the cube and cuboids obtained in Fig. 10.1 to 10.4 to form a solid as shown in Fig. 10.5, which is a cube of side a units.
- Now, remove a cube of side b units from the solid obtained in Fig. 10.5, thus we obtain solid as shown in Fig. 10.6.
For Fig. 10.1, volume of cuboid = (a-b) x a x a = (a-b)a²
For Fig. 10.2, volume of cuboid = (a-b) x a x b = (a-b)ab
For Fig. 10.3, volume of cuboid = (a – b) x b x b = (a – b)b²
For Fig. 10.4, volume of cube =b³
For Fig. 10.5, volume of cube = Sum of volume of all cubes and cuboids
= (a – b)a² + (a – b)ab + (a – b)b² + b³ …..(i)
The cube obtained in Fig. 10.5 has its each side a.
Its volume = (side)³ = a³ …..(ii)
From Eqs. (i) and (ii), we get
a³ = (a – b)a² + (a – b)ab + (a – b)b² + b³ …..(iii)
For Fig. 10.6, volume of solid obtained = a³ – b³
= (a – b)a² + (a – b)ab + (a – b)b² + b³ – b³ [from Eq.(iii)]
= (a – b)a² + (a – b)ab + (a – b)b² = (a-b) (a² +ab + b²)
Therefore, a³-b³ = (a-b) (a²+ab+b²)
Here, volume is in cubic units.
On actual measurement, we get
a = ……. , b = ……. ,
So, a² =…….. , b² = ……. ,
(a- b) = ……. , ab = ……. ,
a³ =…….. , b³ = ……. ,
Hence, a³-b³ = (a-b) (a²+ab+b²).
The algebraic identity a³-b³ = (a-b) (a²+ab+b²) has been verified.
The identity can be used in simplification and factorisation of algebraic expressions.
What is the expanded form of a³ – b³?
Expanded form of a³-b³ = (a-b) (a²+ab+b²).
If (a² + ab + b²) = 0, then what will be the value of a³ – b³?
a³-b³ = (a-b) (a²+ab+b²)
If x = y, then what will be the value of x³ – y³?
Now, x³ – y³ = x³ – x³ = 0
What is degree of the expression y³ – x³ ?
The degree of given expression is 3.
If we replace a by -a and b by -b, then what is the expansion of a³ – b³ ?
a³ + b³ = (-a³)-(-b³)
b³ -a³ =(b-a)(b² +a² +ab)
Verify the algebraic identity x³ – y³ = (x – y) (x² + xy+ y²) by using x =7, y = 5.