Rs aggarwal solutions
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Factorization of Polynomials | Factorization of Polynomials Exercise 3F

Question 29

𝑎^12 − 𝑏^12

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𝑎^12 − 𝑏^12

The given question can be written as

= (𝑎^6)^2 − (𝑏^6)^2

According to the equation

𝑎^2 − 𝑏^2 = (𝑎 + 𝑏)(𝑎 − 𝑏)

So we get,

(𝑎^6)^2 − (𝑏^6)^2

= (𝑎^6 + 𝑏^6)(𝑎^6 − 𝑏^6)

Now we can write it as,

= [(𝑎^2)^3 + (𝑏^2)^3][(𝑎^3)^2 − (𝑏^3)^2]

So according to the equation

𝑎^3 + 𝑏^3 = (𝑎 + 𝑏)(𝑎^2 − 𝑎𝑏 + 𝑏^2)

(𝑎^2 − 𝑏^2) = (𝑎 + 𝑏)(𝑎 − 𝑏)

So we get,

= (𝑎^2 + 𝑏^2)((𝑎^2)^2 − 𝑎^2𝑏^2 + (𝑏^2)^2) (𝑎^3 + 𝑏^3)(𝑎^3 − 𝑏^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,

= (𝑎^2 + 𝑏^2)(𝑎^4 − 𝑎^2𝑏^2 + 𝑏^4) (𝑎 + 𝑏)(𝑎^2 − 𝑎𝑏 + 𝑏^2) (𝑎 − 𝑏)(𝑎^2 + 𝑎𝑏 + 𝑏^2)

= (𝑎 + 𝑏) (𝑎 − 𝑏) (𝑎^2 + 𝑏^2) (𝑎^4 − 𝑎^2𝑏^2 + 𝑏^4) (𝑎^2 − 𝑎𝑏 + 𝑏^2) (𝑎^2 + 𝑎𝑏 + 𝑏^2)

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Question 29

𝑎^12 − 𝑏^12

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

𝑎^12 − 𝑏^12

The given question can be written as

= (𝑎^6)^2 − (𝑏^6)^2

According to the equation

𝑎^2 − 𝑏^2 = (𝑎 + 𝑏)(𝑎 − 𝑏)

So we get,

(𝑎^6)^2 − (𝑏^6)^2

= (𝑎^6 + 𝑏^6)(𝑎^6 − 𝑏^6)

Now we can write it as,

= [(𝑎^2)^3 + (𝑏^2)^3][(𝑎^3)^2 − (𝑏^3)^2]

So according to the equation

𝑎^3 + 𝑏^3 = (𝑎 + 𝑏)(𝑎^2 − 𝑎𝑏 + 𝑏^2)

(𝑎^2 − 𝑏^2) = (𝑎 + 𝑏)(𝑎 − 𝑏)

So we get,

= (𝑎^2 + 𝑏^2)((𝑎^2)^2 − 𝑎^2𝑏^2 + (𝑏^2)^2) (𝑎^3 + 𝑏^3)(𝑎^3 − 𝑏^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,

= (𝑎^2 + 𝑏^2)(𝑎^4 − 𝑎^2𝑏^2 + 𝑏^4) (𝑎 + 𝑏)(𝑎^2 − 𝑎𝑏 + 𝑏^2) (𝑎 − 𝑏)(𝑎^2 + 𝑎𝑏 + 𝑏^2)

= (𝑎 + 𝑏) (𝑎 − 𝑏) (𝑎^2 + 𝑏^2) (𝑎^4 − 𝑎^2𝑏^2 + 𝑏^4) (𝑎^2 − 𝑎𝑏 + 𝑏^2) (𝑎^2 + 𝑎𝑏 + 𝑏^2)

Our top 5% students will be awarded a special scholarship to Lido.

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