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

Question 21 Factorization of Polynomials Exercise 3F

𝑥^6 − 729


𝑥^6 − 729

According to the equation,

𝑎^3 − 𝑏^3 = (𝑎 − 𝑏)(𝑎^2 + 𝑎𝑏 + 𝑏^2)

So we get,

𝑥^6 − 729

= (𝑥^2)^3 − 9^3

Using the equation,

= (𝑥^2 − 9)((𝑥^2)^2 + (𝑥^2)(9) + 9^2)

= (𝑥^2 − 9)(𝑥^4 + 9𝑥^2 + 81)

Based on the equation,

= (𝑥 + 3)(𝑥 − 3)[(𝑥^2 + 9)^2 − (3𝑥)^2]

On further simplification

= (𝑥 + 3)(𝑥 − 3)(𝑥^2 + 9 + 3𝑥)(𝑥^2 + 9 − 3𝑥)

Video transcript
"hello guys welcome back to your homework i'm revant kumar working as a tutor at leader so today we are going to solve and explore one more very interesting question and it's really very easy to solve also so let's see how we can do it here so now we have given here it as x cube minus 729 x power 6 x 6 minus 129 so x power 6 and 729 are also perfect squares only if you are writing them in another way so x square whole cube minus 9 whole cube now they are perfect square only right so now it is in the exact format of a cube minus b cube again so a q minus b q formula is equals to a minus b of a squared plus a b plus b square so just substitute these values here a and b in this then you will be getting the answer as x square minus 9 x square minus 9 of x square whole square plus x square multiplied by 9 plus 9 whole square so that is equals to x square minus 9 of x power 4 plus 9 x square plus 81 so that is our final answer here so i hope you have everything clear like what i've done here 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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