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Rs aggarwal solutions
Grade 9 
Mathematics
Secondary school Mathematics class 9 Rs Aggarwal
Factorization of Polynomials
Factorization of Polynomials Exercise 3F
Question 9
Mathematics
Secondary school Mathematics class 9 Rs Aggarwal
Factorization of Polynomials
Factorization of Polynomials Exercise 3F
Question 9
Factorization of Polynomials | Factorization of Polynomials Exercise 3F
Question 9
CHAPTERS
3. Factorization of Polynomials
4. Linear Equations In Two Variables
6. Introduction To Euclids Geometry
9. Congruence Of Triangles And Inequalities In A Triangle
11. Areas Of Parallelograms And Triangles
14. Areas Of Triangles And Quadrilaterals
15. Volume And Surface Area Of Solids
1 − 27𝑎^3
1 − 27𝑎^3
According to the equation,
𝑎^3 − 𝑏^3 = (𝑎 − 𝑏)(𝑎^2 + 𝑎𝑏 + 𝑏^2)
So we get,
1 − 27𝑎^3
= 1^3 − (3𝑎)^3
Using the formula,
= (1 − 3𝑎)(1^2 + 3𝑎 + (3𝑎)^2)
= (1 − 3𝑎)( 1^2 + 3𝑎 + 9𝑎^2)
= (1 − 3𝑎)( 1 + 3𝑎 + 9𝑎^2)
Question 9
1 − 27𝑎^3
1 − 27𝑎^3
According to the equation,
𝑎^3 − 𝑏^3 = (𝑎 − 𝑏)(𝑎^2 + 𝑎𝑏 + 𝑏^2)
So we get,
1 − 27𝑎^3
= 1^3 − (3𝑎)^3
Using the formula,
= (1 − 3𝑎)(1^2 + 3𝑎 + (3𝑎)^2)
= (1 − 3𝑎)( 1^2 + 3𝑎 + 9𝑎^2)
= (1 − 3𝑎)( 1 + 3𝑎 + 9𝑎^2)
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