Factorization of Polynomials | Factorization of Polynomials Exercise 3F

Question

7𝑎^3 + 56𝑏^3

Solution
Transcript

"hi students welcome to leo learning i hope you are watching all videos so you have solved many more questions before this and you have understood as well so today we have one of the question in the series so the question we have is 7 a cube plus 56 bq now to solve the question first we will take 7 as a common so it will be a cube plus 8 b cube now 7 a cube plus 8 can be written as 2 and b whole cube now here we'll apply the formula which is equals a q plus b cube is equals to a plus b multiplied by a square minus twice minus a b plus b square so 7 will remain as it is it is a in bracket a plus 2 b in for the bracket it's a square minus 2 a b plus 2 b whole square now 7 it's in the first bracket will be as it is 7 a plus b in the second bracket a square minus twice a b plus 4 b square so this way we can simplify this now we'll not simplify anymore after this because doing a multiplication will further go back to the same question so now this will be our final answer so i hope you understood the question we just used one formula aq plus bq for any doubt you can write in the comment section for regular updates do subscribe to the video channel thank you "

Accepted Answer

7𝑎^3 + 56𝑏^3

We can write the given question as,

7𝑎^3 + 56𝑏^3 = 7(𝑎^3 + 8𝑏^3)

According to the equation,

𝑎^3 + 𝑏^3 = (𝑎 + 𝑏)(𝑎^2 − 𝑎𝑏 + 𝑏^2)

So we get,

7(𝑎^3 + 8𝑏^3)

= 7[(𝑎)^3 + (2𝑏)^3]

Using the formula

= 7[(𝑎 + 2𝑏)(𝑎^2 − 𝑎(2𝑏) + (2𝑏)^2]

= 7(𝑎 + 2𝑏)( 𝑎^2 − 2𝑎𝑏 + 4𝑏^2)

Factorization of Polynomials | Factorization of Polynomials Exercise 3F

Chapters in Secondary school Mathematics class 9 Rs Aggarwal

Exercises in Factorization of Polynomials

Question 6