# RS Aggarwal Solutions Class 9 Mathematics Solutions for Factorization of Polynomials Exercise 3F in Chapter 3 - Factorization of Polynomials

Question 30 Factorization of Polynomials Exercise 3F

𝑥^6 − 7𝑥^3 − 8

𝑥^6 − 7𝑥^3 − 8

By substituting 𝑥^3 = 𝑦 in the given equation

We get,

= 𝑦^2 − 7𝑦 − 8

= 𝑦^2 − 8𝑦 + 𝑦 − 8

Taking y as common in the first term and 1 as common in the second term

= 𝑦(𝑦 − 8) + 1(𝑦 − 8)

= (𝑦 + 1)(𝑦 − 8)

Now by replacing y = 𝑥^3

We get,

= (𝑥^3 + 1)(𝑥^3 − 8)

= (𝑥^3 + 1^3)(𝑥^3 − 2^3)

According to the equation

𝑎^3 + 𝑏^3 = (𝑎 + 𝑏)(𝑎^2 − 𝑎𝑏 + 𝑏^2)

𝑎^3 − 𝑏^3 = (𝑎 − 𝑏)(𝑎^2 + 𝑎𝑏 + 𝑏^2)

We get,

= (𝑥 + 1)(𝑥^2 − 𝑥 + 1^2) (𝑥 − 2)(𝑥^2 + 2𝑥 + 2^2)

On further simplification

= (𝑥 + 1)(𝑥^2 − 𝑥 + 1) (𝑥 − 2)(𝑥^2 + 2𝑥 + 4)

= (𝑥 + 1) (𝑥 − 2) (𝑥^2 − 𝑥 + 1) (𝑥^2 + 2𝑥 + 4)

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