Question

guys is my answer right? thankyou

(x^3-27)/(x^3+64) times (x+4)/(x2+3x+9)

=(x-3)(x^2+3x+9)/(x+4)(x^2-4x+16) times (x+4)/(x^2+3x+9

=(x+4)/x^23x+9

is this right?? thankyou guys ^^ takecare

Answer 1

take care? where you going

Answer 2

what did u do after this step ?

(x-3)(x^2+3x+9)/(x+4)(x^2-4x+16) times (x+4)/(x^2+3x+9

x+4 gets cancelled, isn't it ?

Answer 3

and (x^2+3x+9)

Answer 4

\[x-3 \]

Answer 5

so what remains is (x-3) / (x^2-4x+16)

Answer 6

@hartnn yes i cancelled it

Answer 7

@perl LOL.

Answer 8

:)

Answer 9

@hartnn yes. is it wrong?

@curiousshubham . is the ans x-3?

Answer 10

@xtel yep!

Answer 11

x+4 gets cancelled ,
your answer --->(x+4)/x^2+3x+9
are the terms that actually gets cancelled
so what remains is
(x-3) / (x^2-4x+16)

Answer 12

@curiousshubham where did (x^2-4x+16) go ?

Answer 13

@hartnn is cancelling the nos. is the proper way ? thankyou

Answer 14

@xtel \[\frac{ x^{3}-27 }{ x^{3}+64 }\times \frac{ x+4 }{ x^{2}+3x+9 }\] This is your question?

Answer 15

@curiousshubham yes, thats it

Answer 16

Sorry @hartnn I left \[x^{2}-4x+16\] @xtel the solution is \[\frac{ x-3 }{ x^{2}-4x+16 }\]

Answer 17

no problem.

Answer 18

@curiousshubham and @hartnn thankyou guys :))

Answer 19

welcome :)

Answer 20

you are welcomed...

Answer 21

:)