I'm stuck on a question on Binomial Expressions.. any help? c:

[Lexadis]

Solved://I'm stuck on a question on Binomial Expressions.. any help? c:

So currently we are studying expressions based on (x+b)³ and (x-b)^3.
.
In this aspect, there's this question..
Q. Obtain the expression of (x-y)³, by substituting (-y) for y in the expansion of (x+y)³.

So.. I tried several ways, and I hope the second method is correct, but I just wanted to be sure ;D

Method I
I tried expanding it instead of using the formula, and got stuck in half :/
(x+y)³ = (x+y) (x+y)²
= (x-y) (x-y)²
= (x-y) (x²-2xy+y²)
= x(x²-2xy+y²) - y(x²-2xy+y²)
= x³ - 2x²y + xy² - x²y + 2xy² - y³
= x³ - 3x²y + 3xy² - y³
=?
I know that this is the equation which finally builds upto (x-y)³, but I just don't know how to reconstruct it back :sad:
Method II
And here again I tried expanding directly through the formula, and got stuck in the same place:
(x+y)³ = x³ + 3x²y + 3xy² + y³
= x³ + 3x²(-y) + 3x(-y)² + (-y³)
= x³ - 3x²y + 3xy² - y³
= ?

So.. I tried solving after this, but it got me wrong. Here's my result:
= x³ - 3x²y + 3xy² - y³

= x² (x-3y) - y² (y-3x)
= (x-y)²(x-3y)(y-3x)
= (x-y)(x+y)(x-3y)(y-3x)
= ?

Any help appreciated. Thank you c:

 

[DrPhil]

So currently we are studying expressions based on (x+b)³ and (x-b)^3.
.
In this aspect, there's this question..
Q. Obtain the expression of (x-y)³, by substituting (-y) for y in the expansion of (x+y)³.

So.. I tried several ways, and I hope the second method is correct, but I just wanted to be sure ;D

Method I
I tried expanding it instead of using the formula, and got stuck in half :/
(x+y)³ = (x+y) (x+y)²
(x-y)³ = (x-y) (x-y)²
= (x-y) (x²-2xy+y²)
= x(x²-2xy+y²) - y(x²-2xy+y²)
= x³ - 2x²y + xy² - x²y + 2xy² - y³
= x³ - 3x²y + 3xy² - y³
= the correct answer!
I know that this is the equation which finally builds upto (x-y)³, but I just don't know how to reconstruct it back :sad:
Method II
And here again I tried expanding directly through the formula, and got stuck in the same place:
(x+y)³ = x³ + 3x²y + 3xy² + y³
(x-y)³ = x³ + 3x²(-y) + 3x(-y)² + (-y³)
= x³ - 3x²y + 3xy² - y³
= the correct answer!

So.. I tried solving after this, but it got me wrong. Here's my result:
= x³ - 3x²y + 3xy² - y³

= x² (x-3y) - y² (y-3x)
= (x-y)²(x-3y)(y-3x) NO
= (x-y)(x+y)(x-3y)(y-3x)
= ?

Any help appreciated. Thank you c:

You did it right two times! If you want to factor the result, try dividing by (x-y).

 

[Lexadis]

You did it right two times! If you want to factor the result, try dividing by (x-y).

I'm still finding it unable to do it :/ The fact is, we have to get (x-y)³ from x³ - 3x²y + 3xy² - y³.
So I tried factorizing, as you stated, by (x-y) and I just can't do it >.> Anyways, this was what I tried:
x³ - 3x²y + 3xy² - y³
=x²(x-3y) + y² (3x-y)
=(x+y)²(x-3y)(3x-y)
=(x+y)²(x-y)(3x-3)
=?

I'm not sure of how to facorize them by (x-y) either. Any help? Thank you very much c:

---edit---
I tried another method, but I didn't get the answer I needed. But still, I got a completely different one. So here it goes:
x³ - 3x²y + 3xy² - y³
=x²(x-3y) + y² (3x-y)
=(x+y)²(x-3y)(3x-y)
= x² + 2xy + y² + 3x² - 10xy + 3y² [I expanded (x+y)², as well as (x-y)(3x-3)]
= 4x² - 8xy + 4y²
= 4 (x² - 2xy + y²)
= 4 [x(x-y)-y(x-y)]
= 4(x-y)(x-y)
=4 (x-y)²
= ?
I have to get (x-y)³ as the end result, but I seem to have ended up getting something else Any help as to where I have gone wrong is appreciated, Thank you. c:

 

[Deleted member 4993]

The fact is, we have to get (x-y)³ from x³ - 3x²y + 3xy² - y³.

x³ - 3x²y + 3xy² - y³

= x³ - 2x²y + xy² -x²y + 2xy² - y³

= x(x² - 2xy + y²) - y(x² - 2xy + y²)

Can you go from here....

 

[DrPhil]

I'm still finding it unable to do it :/ The fact is, we have to get (x-y)³ from x³ - 3x²y + 3xy² - y³.

Having found the expansion of (x-3)³ by substituting -y for y in the expansion of (x+y)³, namely doing your Method II, write down all the steps from your Method I, IN REVERSE ORDER.

(x-y)³ = (x-y) (x-y)²
= (x-y) (x²-2xy+y²)
= x(x²-2xy+y²) - y(x²-2xy+y²)
= x³ - 2x²y + xy² - x²y + 2xy² - y³
= x³ - 3x²y + 3xy² - y³

x³ - 3x²y + 3xy² - y³
= x³ - 2x²y + xy² - x²y + 2xy² - y³
= x(x²-2xy+y²) - y(x²-2xy+y²)
= (x-y) (x²-2xy+y²)
= (x-y) (x-y)²
= (x-y)³

 

[Lexadis]

Thank you for both Subhotosh Khan and DrPhil for helping me