Sum/Difference of Cubes in Brackets

Dalmain

New member
Joined
Apr 15, 2021
Messages
7
Good Day All
Hope you are keeping well.

I understand the basic concept of sum/difference of cubes where a^3+b^3 = (a+b)(a^2-ab+b^2) and a^3-b^3 = (a-b)(a^2+ab+b^2)
But I've been thrown a curveball of factorizing

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

now I've been given the solution of

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

without the in-between steps. I don't understand how to get to this answer. Any assistance would be appreciated.

Thank you
Dalmain
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
11,540
I understand the basic concept of sum/difference of cubes where a^3+b^3 = (a+b)(a^2-ab+b^2) and a^3-b^3 = (a-b)(a^2+ab+b^2)
But I've been thrown a curveball of factorizing

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

now I've been given the solution of

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

without the in-between steps. I don't understand how to get to this answer. Any assistance would be appreciated.

Thank you
Dalmain
Just replace a with (x-y) and b with (x+y) in the formula. Then expand (simplify) each factor.

For example, the first factor, (a-b), becomes ((x-y) - (x+y)), which simplifies to (x-y-x-y) = (-2y), just as they show.

Do the same for the second factor, (a^2+ab+b^2), and show us your work if you don't get it down to (3x^2+y^2).
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
11,039
I understand the basic concept of sum/difference of cubes where a^3+b^3 = (a+b)(a^2-ab+b^2) and a^3-b^3 = (a-b)(a^2+ab+b^2) But I've been thrown a curveball of factorizing
(x-y)^3 - (x+y)^3 now I've been given the solution of -2y(3x^2+y^2).
\((x-y)^3=x^3-3x^2y+3xy^2-y^3~\&~(x+y)^3=x^3+3x^2y+3xy^2+y^3\)
Surely you can subtract?
 

Dalmain

New member
Joined
Apr 15, 2021
Messages
7
Good Morning

Thank you both for your replies, I've got it now.
It's been a long time since I've had to do math like this and I will probably be back for more assistance.
I'm afraid you are going to be subjected to my ignorance for a little longer.

Thank you again
 
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