Question

Factor (3x+2y)^3+(y-3)^3

Answer 1

And i already know the formula, i just cant seem to get the correct answer

Answer 2

Ok
\[a^3+b^3=(a+b)(a^2-ab+b^2)\]
So what is a and b here?
a=(3x+2y)
b=(y-3)
Plug in and have fun

Answer 3

like i said, i plug it in, and still get the wrong answer, i end up getting 9x^2+12xy+4y^2-3xy-9x-6y+2y^2+y^2-6y+9

Answer 4

Well it says factor

Answer 5

You didn't leave it in factored form

Answer 6

(3x+2y+y-3)(9x^2+12xy+4y^2- 3xy-9x-6y+2y^2+y^2 -6y +9)

Answer 7

(3x+2y)(y-3)
3x(y-3)+2y(y-3)
3xy-9x+2y^2-6y
---
-(3x+2y)(y-3)
-(3xy-9x+2y^2-6y)
-3xy+9x-2y^2+6y

Answer 8

You didn't distribute that -1 correctly

Answer 9

what was the (3x+2y)(y-3), was that for the second part of the equation?

Answer 10

The -ab part

Answer 11

how did this (3x+2y)(y-3) go to this 3x(y-3)+2y(y-3)

Answer 12

\[(a+b)(c+d)=a(c+d)+b(c+d)\]

Answer 13

\[(a+b+e)(c+d)=a(c+d)+b(c+d)+e(c+d)\]

Answer 14

\[(a+b+e+j)(c+d+e)=a(c+d+e)+b(c+d+e)+e(c+d+e)+j(c+d+e)\]

Answer 15

so what factored out of the (3x+2y) to get (y-3)

Answer 16

?

Answer 17

I was just saying you didn't distribute the -1 in front of ab correctly

Answer 18

do you understand what i'm saying?

Answer 19

\[a^3+b^3=(a+b)(a^2-ab+b^2) \]
a=3x+2y
b=y-3

Answer 20

See that our formula requires -ab?

Answer 21

\[-ab=-(3x+2y)(y-3)\]
\[=-[3x(y-3)+2y(y-3)]\]
\[=-[3xy-9x+2y^2-6y]\]
\[=-3xy+9x-2y^2+6y\]

Answer 22

You have the a^2 and b^2 part right
just the -ab part was wrong

Answer 23

\[(3x+2y+y-3)(9x^2+12xy+4y^2-3xy+9x-2y^2+6y+y^2-6x+9)\]
Simplify the insiders of each ( ) and you are done

Answer 24

(3x+3y-3)(9x^2-9xy+3y^2-3x+6y+9)

Answer 25

I see 9x-6x which is 3x not -3x

Answer 26

and 12xy-3xy=9xy not -9xy

Answer 27

Ok I think that looks right now you could actually factor more lol

Answer 28

3(x+y-1)3(3x^2+3xy+y^2+x+2y+3)?

Answer 29

Right and then 3*3=9

Answer 30

so then 9(x+y-1)(3x^2+3xy+y^2+x+2y+3), so is that the final answer?

Answer 31

Looks good to me

Answer 32

I don't think there is suppose to be a 2y though

Answer 33

I messed up somewhere, cuz the final answer in the book doesn't have 2y

Answer 34

yeah the 6y-6y=0
so there is no term in the ( ) with just variable part y

Answer 35

\[(3x+2y+y-3)(9x^2+12xy+4y^2-3xy+9x-2y^2+6y+y^2-6y+9) \]
see in the second parathesis 6y-6y=0

Answer 36

and also there is 9x

Answer 37

\[(3x+3y-3)(9x^2+9xy+6y^2+9x+9)\]

Answer 38

Now factor the 3's out from each ( )

Answer 39

oh okay i see it