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

Question 20 Factorization of Polynomials Exercise 3F

1029 − 3𝑥^3


1029 − 3𝑥^3

We can write the given question as,

1029 − 3𝑥^3 = 3(343 − 𝑥^3)

According to the equation,

𝑎^3 − 𝑏^3 = (𝑎 − 𝑏)(𝑎^2 + 𝑎𝑏 + 𝑏^2)

So we get,

3(343 − 𝑥^3)

= 3[7^3 − 𝑥^3]

Based on the formula,

= 3[(7 − 𝑥)(7^2 + 7𝑥 + 𝑥^2)]

= 3(7 − 𝑥)(49 + 7𝑥 + 𝑥^2)

Video transcript
"hello guys welcome back to your homework i'm ravanth kumar working as a tutor at leader so today we are going to solve one very interesting question and it's really very easy so how we can do it so let's see here 1029 minus 3 x cube so this is the question i have given here so this is 1 0 2 9 and 3 x cube are not the perfect cubes so we need to take some common here so that is if you are taking 3 as common here it will be remaining with 343 minus x cube right so now 343 and x cube are the perfect cubes so now 343 can be written in the form of x whole cube minus x whole cube right so your three will be the outside then it is in the format of a cube minus b cube so that is equals to a minus b of a square plus a b plus b square right so that is equals to here uh just substitute each and everything values here a and b so that is 7 minus x of 7 minus x of 7 squared plus a b that is 7x plus and your b square that is x square so that will be resulting of 3 of 7 minus x so whole of that is 7 square means 49 plus 7x plus x square so that is only written here so i hope it is everything clear to everyone if you have any doubts please comment below don't forget to subscribe to this channel thank you so much for watching this stay tuned thank you "
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