Prove x^3 - y^3 = (x - y)(x^2 +xy + y^2)

  • Thread starter zeion
  • Start date
In summary, the original poster is trying to find a factor of (x3 - y3) but isn't sure how to do it. They mention that they need to use some basic properties and then say they're not supposed to expand the right side. They also mention that if the RHS wasn't given, they could find the factor using addition and multiplication.
  • #1
zeion
466
1

Homework Statement



Prove [tex] x^3 - y^3 = (x - y)(x^2 +xy + y^2) [/tex]

Homework Equations





The Attempt at a Solution



Not sure.. I know I'm not supposed to expand the right side. lol.
Need to use some basic properties?
 
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  • #2
Why can't you expand the right side?
 
  • #3
Do you know how to determine the roots of

[tex]x^3-1=0?[/tex]

If so, can you use them to write down the roots of

[tex]x^3-y^3=0?[/tex]
 
  • #4
Uhh..

[tex] x^3 - 1 = 0 \Rightarrow x^3 = 1 \Rightarrow x = 1 [/tex] ?
 
  • #5
ok you obviously don't know complex numbers then, again, why can't you just expand the right side? Why did you say "I know I'm not supposed to expand the right side"?
 
  • #6
zeion said:
Uhh..

[tex] x^3 - 1 = 0 \Rightarrow x^3 = 1 \Rightarrow x = 1 [/tex] ?

That's only one root. Since it's a cubic equation there are three roots. If we ignore that for the moment, can you use the root x = 1 to write down a factor of (x3 - y3)? The remaining factor will be quadratic and there's a couple of ways you might find the coefficients.
 
  • #7
Okay so one root is x = y
So (x - y) is a root.
 
  • #8
zeion said:
Okay so one root is x = y
So (x - y) is a root.

(x - y) is a factor. The remaining factors give a quadratic form, any ideas on how to determine the coefficients?
 
  • #9
Um long division?
 
  • #10
An easier way to do this would be to expand the right hand side as mentallic suggested.
And i presume that since you don't know complex numbers this is probably the easiest way.
:smile:
 
  • #11
zeion said:
Um long division?

Sure, if you know how to do that, it's a good way.
 
  • #12
But is there a way to prove this using the properties of addition and multiplication?
 
  • #13
zeion said:
But is there a way to prove this using the properties of addition and multiplication?

You are using addition and multiplication. If the RHS wasn't given and you were asked to factor x3 - y3, then you would first find the roots. If you didn't know about roots of unity, you'd only be able to find the real root. This gives one factor, if you know how to divide polynomials, you can find the other factor by division.
 
  • #14
zeion said:
But is there a way to prove this using the properties of addition and multiplication?

Uhh, yeah! There is more than one way to prove something, but the easiest would still be to expand the right side :-p

Try it geometrically. First as a warmup, see if you can prove a2-b2=(a-b)(a+b) by drawing a big square of side length a with a small square of side length b inside it (the smaller square should share corners with the larger square).
 
  • #15
Mentallic said:
Uhh, yeah! There is more than one way to prove something, but the easiest would still be to expand the right side :-p

Try it geometrically. First as a warmup, see if you can prove a2-b2=(a-b)(a+b) by drawing a big square of side length a with a small square of side length b inside it (the smaller square should share corners with the larger square).

What if the RHS wasn't given? Give the guy credit for wanting to learn how to actually learn how to compute something rather than just verify an answer.
 
  • #16
The RHS can be shown by some simple factorizing of the form ab+ac=a(b+c). Try it yourself.
 
  • #17
zeion said:

Homework Statement



Prove [tex] x^3 - y^3 = (x - y)(x^2 +xy + y^2) [/tex]

Homework Equations





The Attempt at a Solution



Not sure.. I know I'm not supposed to expand the right side. lol.
Need to use some basic properties?

This isn't hard if you expand the RHS. [tex]x^3 + x^2y + xy^2 - x^2y - xy^2 - y^3[/tex] and the rest easy...
 
  • #18
fzero said:
What if the RHS wasn't given? Give the guy credit for wanting to learn how to actually learn how to compute something rather than just verify an answer.
But the RHS was given. First answer the question asked, then add whatever other information you want. The difficulty here was clearly the statement "I know I'm not supposed to expand the right side. lol."
 
  • #19
HallsofIvy said:
But the RHS was given. First answer the question asked, then add whatever other information you want. The difficulty here was clearly the statement "I know I'm not supposed to expand the right side. lol."

It sure was Hall, but I had a feeling that original poster didn't know that

if you have [tex](a+b)(a+b)= a^2 + ab + ab + b^2[/tex]

It hard to be young these days!
 
  • #20
No, I think he knows how to expand since he's dealing with the difference of two cubes. He's more interested in how it works (besides applying the usual algebraic techniques) like fzero said.
 
  • #21
For example in the last question it asked me to prove:

[tex] x^2 + y^2 = (x + y)(x - y) [/tex]

So I did something like

[tex] x^2 + y^2 \Rightarrow x^2 + xy - xy + y^2 \Rightarrow x(x + y) - y(x + y) \Rightarrow (x + y)(x - y) [/tex]

So I was just wondering how to do something similar with this one.
 
  • #22
zeion said:
For example in the last question it asked me to prove:

[tex] x^2 + y^2 = (x + y)(x - y) [/tex]
This is not true. However, x2 - y2 = (x - y)(x + y)
zeion said:
So I did something like

[tex] x^2 + y^2 \Rightarrow x^2 + xy - xy + y^2 \Rightarrow x(x + y) - y(x + y) \Rightarrow (x + y)(x - y) [/tex]
Use = for expressions that are equal, not [itex]\Rightarrow[/itex].
x2 + xy - xy + y2 = x(x + y) - y(x - y). Note that this is not equal to (x - y)(x + y).
zeion said:
So I was just wondering how to do something similar with this one.
 
  • #23
zeion, you mean x2-y2=(x+y)(x-y)

Well, working backwards from (x+y)(x-y) you get x(x-y)+y(x-y)=x2-xy+xy-y2

For the difference of two cubes, x3-y3=(x-y)(x2+xy+y2) so again working backwards you get x(x2+xy+y2)-y(x2+xy+y2)

expanding,

x3+x2y+xy2-x2y-xy2-y3

So to factorize x3-y3 you would need to know that you have to add and subtract x2y and xy2 from the expression and factorize from there.

The geometric technique I mentioned helps you do this without having to guess, it quickly shows you what you must add and subtract in order to factorize the expression.
 
  • #24
Mentallic said:
Uhh, yeah! There is more than one way to prove something, but the easiest would still be to expand the right side :-p

Try it geometrically. First as a warmup, see if you can prove a2-b2=(a-b)(a+b) by drawing a big square of side length a with a small square of side length b inside it (the smaller square should share corners with the larger square).

Ok I tried this on the cube:

So I want to take out a small cube b^3 from big cube a^3
I split the remaining cube into rectangle cubes and added the area:

[tex] a^3 - b^3 = (a - b)(a)(b) + (a - b)(b)(b) + (a - b)(a)(a) = (a - b)(a^2 + ab + b^2) [/tex]

Is there a way to extend this to work for more than 3D?
What if a < b? lol?
 
Last edited:
  • #25
I thought about that question too. It is possible to extend this idea by evaluating the properties of a 4-d shape but we can't draw it on paper in any reasonable way, so it would be venturing into a theoretical world again which won't be any more of a useful tool to prove it than expanding the factorized expression would be.

By the way, if it asked you in a book to prove the difference of two cubes, almost surely it expected you to just expand the right side, or possibly using polynomial division.
 
  • #26
Here is a method that doesn't involve manipulating the right hand side, which is more I think of what we're looking for:

Artificially create a factor (x - y) and add canceling terms at the end:

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

With the final two terms cancelling the terms on the right hand side that don't appear on the left hand side.

Then pull out another ( x - y ) on the final two terms:

x^2 * y - y^2 * x = ( x - y )( x * y )

Put it together and you're done!
 
  • #27
creillyucla said:
Here is a method that doesn't involve manipulating the right hand side, which is more I think of what we're looking for:

Artificially create a factor (x - y) and add canceling terms at the end:

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

With the final two terms cancelling the terms on the right hand side that don't appear on the left hand side.

Then pull out another ( x - y ) on the final two terms:

x^2 * y - y^2 * x = ( x - y )( x * y )

Put it together and you're done!

This is an intuitively based response and I don't see it working for degrees higher than 3. You know where it's meant to lead to, so what if we asked you to try do this for x7+y7? Without knowing the pattern or ever having seen the second factor, it won't be an easy task to do. Polynomial division however is a mathematical process which doesn't require intuitive guesses.
 

Related to Prove x^3 - y^3 = (x - y)(x^2 +xy + y^2)

What is the equation x^3 - y^3 = (x - y)(x^2 +xy + y^2) used for?

The equation x^3 - y^3 = (x - y)(x^2 +xy + y^2) is used to factorize a cubic polynomial into two binomials.

How do you prove x^3 - y^3 = (x - y)(x^2 +xy + y^2)?

To prove x^3 - y^3 = (x - y)(x^2 +xy + y^2), you can use the distributive property to expand the right side of the equation and then simplify it to match the left side.

What is the significance of the equation x^3 - y^3 = (x - y)(x^2 +xy + y^2)?

The equation x^3 - y^3 = (x - y)(x^2 +xy + y^2) is a special case of the difference of cubes formula, which is commonly used in algebraic manipulation and factoring.

How can I use x^3 - y^3 = (x - y)(x^2 +xy + y^2) to solve equations?

You can use the equation x^3 - y^3 = (x - y)(x^2 +xy + y^2) to factorize cubic equations and solve for the roots of the equation.

Can x^3 - y^3 = (x - y)(x^2 +xy + y^2) be applied to real-life situations?

Yes, the equation x^3 - y^3 = (x - y)(x^2 +xy + y^2) can be used in real-life situations such as calculating the volume of a cube or factoring a polynomial in engineering or physics problems.

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