# Thread: The Sum and difference of cubes

1. ## The Sum and difference of cubes

Hello,
Need help in these problems

1-(x+y)^3

(x+y)^3+1

1-a^6x^6

I am having some trouble figuring these out, if you can help me please, thanks.

If you can key it out fully would be great to help me understand.
Thank you
Joanne

Hello,
Need help in these problems

1-(x+y)^3
you have $1^3 - (x + y)^3$

apply the difference of two cubes formula: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

here $a = 1$, $b = x + y$

try it

(x+y)^3+1
formula: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

here, $a = x + y$, $b = 1$

1-a^6x^6
write it as $1 - [(ax)^3]^2$.

applying the difference of two squares formula, we get: $[1 + (ax)^3][1 - (ax)^3]$

now, apply the sum of two cubes formula to the first factor, and the difference of two cubes formula to the last factor

Hello,
Need help in these problems

1-(x+y)^3

(x+y)^3+1

1-a^6x^6

I am having some trouble figuring these out, if you can help me please, thanks.

If you can key it out fully would be great to help me understand.
Thank you
Joanne
note the factoring patterns for the sum and difference of two cubes ...

$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$

$a^3 - b^3 = (a-b)(a^2 + ab + b^2)$

for the first problem ...

$a$ is $1$

$b$ is $(x+y)$

$1^3 - (x+y)^3 = [1 - (x+y)][1^2 + 1(x+y) + (x+y)^2]$

4. use binomial theorem
it is ussually in the book cover or listed in the reference

$(x+y)^3 = x^3+3x^2y+3xy^2+y^3$

5. ## Hello Again and thanks for the reply.

1-(x+y)^3 = (1-x-y)(1+x+y+(x+y)^2 which I got now, thanks.

(x+y)^3+1 = I can't find the full answer to this, sorry.

1-a^6x^6 = (1-ax)(1+ax)(1+ax+a^2x^2)(1-ax+a^2x^2)
This one I am still stuck on doing. I must be doing something wrong, and can't figure it out.
Can someone completely display this answer, thanks.
Joanne

(x+y)^3+1 = I can't find the full answer to this, sorry.
$[(x+y)+1][(x+y)^2 - (x+y)(1) + 1^2]$

clean it up.

for $1-a^6x^6$ , try factoring ...

$1^3 - (a^2x^2)^3$

7. theorum

$(x+y)^3 = x^3+3x^2y+3xy^2+y^3$

so just add 1 to it

$(x+y)^3+1 = x^3+3x^2y+3xy^2+y^3+1$

(x+y)^3+1 = I can't find the full answer to this, sorry.
skeeter dealt with this

1-a^6x^6 = (1-ax)(1+ax)(1+ax+a^2x^2)(1-ax+a^2x^2)
This one I am still stuck on doing. I must be doing something wrong, and can't figure it out.
Can someone completely display this answer, thanks.
Joanne
what you have here is correct

9. Originally Posted by bigwave
use binomial theorem
it is ussually in the book cover or listed in the reference

$(x+y)^3 = x^3+3x^2y+3xy^2+y^3$
Originally Posted by bigwave
theorum

$(x+y)^3 = x^3+3x^2y+3xy^2+y^3$

so just add 1 to it

$(x+y)^3+1 = x^3+3x^2y+3xy^2+y^3+1$
well, the OP was a little vague, but judging from the title, i think the idea here is to factor, not expand.

10. 1-a^6x^6 = (1-ax)(1+ax)(1+ax+a^2x^2)(1-ax+a^2x^2)

That is the answer and I am trying this and still not getting anywhere.
I get (1-a^2x^2) (1^2+ax+a^4x^4)
So then you get(1-ax)(1+ax) but stuck on this if it's suppose to look like the answer above.
thks
Jo

11. ## Skeeter

The reply you have above saying I have it right. That is the answer in my work book. I am stuck on getting the last part of the problem as I noted above.
Jo

if you apply the difference of two cubes first (which makes the problem harder) you would get $[1 - (ax)^2][1 + (ax)^2 + (ax)^4]$
you would then need to factor $1 - (ax)^2$ using the difference of two squares, and to factor $1 + (ax)^2 + (ax)^4$, you would need to write it as $(ax)^4 + 2(ax)^2 + 1 - (ax)^2 = [(ax)^2 + 1]^2 - (ax)^2$ and use the difference of two sqaures again. See? too much work. review my first post and follow the suggestion there