(x-y)3 + (y-z)3 + (z-y)3 = 3(x-y)(y-z)(z-x)

[lovatto]

Homework Statement

Hello

I am starting to learn mathematics, currently working through

Fundamentals of University Mathematics (Woodhead Publishing in Mathematics)
Colin M. McGregor (Author), John Nimmo (Author), Wilson W. Stothers (Author)

http://www.amazon.com/gp/product/0...ls_o01_s00_i00&tag=

I have come across a stumbling block with Exercise 1.5.

The exercise is to:

Show that
(x-y)^3 +(y-z)3 + (z-x)^3 = 3(x-y)(y-z)(z-x)


That is it no constraints etc. It mentions: "This can be done by expanding out the brackets, but there is a more elegant solution."

Homework Equations

The Attempt at a Solution

First of all this only seems to hold in special cases as I have substituted random values for x,y and z and they do not agree.

It seems more like this should be an inequality

(x-y)^3 +(y-z)3 + (z-x)^3 >= 3(x-y)(y-z)(z-x)

But that is not the question set.

Please note that this is the first chapter and all that has been covered is basic number theory, rational powers, inequalities and divisibility. I am assuming those are the only tools I have at my disposal, I have not been introduced to any identities etc.

I decided to expand out the brackets but seem to be stuck attempting that method also

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

x3-y3-xy(1+2x-2y-y) + y3-z3-yz(1+2y-2z-z) + z3-x3-zx(1+2z-2x-x) = 3(x-y)(y-z)(z-x)

all the cubes then cancel

-xy(1+2x-2y-y) -yz(1+2y-2z-z) -zx(1+2z-2x-x) = 3(x-y)(y-z)(z-x)

Then I start to struggle to see where I can go to make the two equivalent.

What I really want to know is this (unmentioned in the solutions section) "elegant solution".

The only thing I am aware of is this identity (this is not mentioned in the textbook so I cannot use this to help me)

a3+b3+c3-3abc = (a+b+c)(a2+b2+c2-ab-bc-ac)

And for this identity if a+b+c=0 then obviously a3+b3+c3-3abc=0

If i substitute (x-y)(y-z)(z-x) for a,b and c respectively

a+b+c=x-y+y-z+z-x

The x's y's and z's cancel giving

a+b+c=z-y+y-z+z-x=0

I think that would seem to be on the right lines but this is not mentioned anywhere in the textbook.

Basically I am all muddled up.

 

[oli4]

Hi, I think you must have read it wrong
your left hand side of the equation looks wrong
it should be (x-y)³+(y-z)³+(z-x)³
try expanding this one, it should work fine.
What you wrote first, since (-a)³=-(a³), could be reduced to just (x-z)³.

Cheers...

 

[Ray Vickson]

Homework Statement

Hello

I am starting to learn mathematics, currently working through

Fundamentals of University Mathematics (Woodhead Publishing in Mathematics)
Colin M. McGregor (Author), John Nimmo (Author), Wilson W. Stothers (Author)

http://www.amazon.com/gp/product/0...ls_o01_s00_i00&tag=

I have come across a stumbling block with Exercise 1.5.

The exercise is to:

Show that
(x-y)3 + (y-z)3 + (z-y)3 = 3(x-y)(y-z)(z-x)


That is it no constraints etc. It mentions: "This can be done by expanding out the brackets, but there is a more elegant solution."

Homework Equations

The Attempt at a Solution

First of all this only seems to hold in special cases as I have substituted random values for x,y and z and they do not agree.

It seems more like this should be an inequality

(x-y)3 + (y-z)3 + (z-y)3 >= 3(x-y)(y-z)(z-x)

But that is not the question set.

Please note that this is the first chapter and all that has been covered is basic number theory, rational powers, inequalities and divisibility. I am assuming those are the only tools I have at my disposal, I have not been introduced to any identities etc.

I decided to expand out the brackets but seem to be stuck attempting that method also

(x-y)3 + (y-z)3 + (z-y)3 = 3(x-y)(y-z)(z-x)

x3-y3-xy(1+2x-2y-y) + y3-z3-yz(1+2y-2z-z) + z3-x3-zx(1+2z-2x-x) = 3(x-y)(y-z)(z-x)

all the cubes then cancel

-xy(1+2x-2y-y) -yz(1+2y-2z-z) -zx(1+2z-2x-x) = 3(x-y)(y-z)(z-x)

Then I start to struggle to see where I can go to make the two equivalent.

What I really want to know is this (unmentioned in the solutions section) "elegant solution".

The only thing I am aware of is this identity (this is not mentioned in the textbook so I cannot use this to help me)

a3+b3+c3-3abc = (a+b+c)(a2+b2+c2-ab-bc-ac)

And for this identity if a+b+c=0 then obviously a3+b3+c3-3abc=0

If i substitute (x-y)(y-z)(z-x) for a,b and c respectively

a+b+c=x-y+y-z+z-x

The x's y's and z's cancel giving

a+b+c=z-y+y-z+z-x=0

I think that would seem to be on the right lines but this is not mentioned anywhere in the textbook.

Basically I am all muddled up.

Please use proper symbols: either use (x-y)^3 [NOT (x-y)3] or else use the "X²" button at the top of the input panel, which would give you (x-y)³. Anyway, your question is not correct; I think you mean (x-y)³+(y-z)³+(z-x)³ on the left-hand-side. If you copied the question exactly from the book, then the book is in error.

RGV

 

[lovatto]

Hi, I think you must have read it wrong
your left hand side of the equation looks wrong
it should be (x-y)³+(y-z)³+(z-x)³
try expanding this one, it should work fine.
What you wrote first, since (-a)³=-(a³), could be reduced to just (x-z)³.

Cheers...

Please use proper symbols: either use (x-y)^3 [NOT (x-y)3] or else use the "X²" button at the top of the input panel, which would give you (x-y)³. Anyway, your question is not correct; I think you mean (x-y)³+(y-z)³+(z-x)³ on the left-hand-side. If you copied the question exactly from the book, then the book is in error.

RGV

Hi

Thanks for your replys but that was simply a typo.

I was able to expand it out and it did cancel to 0

-3 x^2 y+3 x^2 z+3 x y^2-3 x z^2-y^3+3 y+z^3-3 z = -3 x^2 y+3 x^2 z+3 x y^2-3 x z^2-3 y^2 z+3 y z^2

What I would like to know is the elegant solution?

 

[Ray Vickson]

Hi

Thanks for your replys but that was simply a typo.

I was able to expand it out and it did cancel to 0

-3 x^2 y+3 x^2 z+3 x y^2-3 x z^2-y^3+3 y+z^3-3 z = -3 x^2 y+3 x^2 z+3 x y^2-3 x z^2-3 y^2 z+3 y z^2

What I would like to know is the elegant solution?

It depends on how much you know already. It is quite easy if you know about the connection between factors and roots.

First, notice that (without doing the complete expansion) you can see that the cubed terms cancel; for example, the x^3 from the first term cancels the -x^3 from the third term, etc. So, the left-hand-side is a quadratic in x, y and z, in the sense their highest powers are 2. Look at the left-hand-side L as a function of x; it is quadratic in x and vanishes when x = y or x = z [because when x = y it is 0^3 + (x-z)^3 + (z-x)^3 = 0, using x instead of y in the second term]. So, the quadratic L factors as L = b*(x-y)(x-z), where b does not depend on x. Note that we can write this as L = c*(x-y)*(z-x), where c = -b; the c-form is closest to what we want, so let's use it. Similarly, as a function of y, L vanishes when y = x or y = z, so (y-x) and (y-z) are both factors of L; that is, our previous c must have y-z as a factor. Finally, we get L = d*(x-y)*(y-z)*(z-x), where d is a constant. Now we can determine d by substituting some convenient values such as z=0, y=1, z=2 into both sides.

RGV