Proving Algebraic Identity: How to Simplify Complex Equations

[NotaMathPerson]

Hello everyone. I need help on this one

Prove that $ \left((a-b)^2+(b-c)^2+(c-a)^2\right)^2=2\left((a-b)^4+(b-c)^4+(c-a)^4\right)$I noticed that the leftside of the eqn when expanded would be like
$(X^2+Y^2+Z^2+2XY+2XZ+2YZ)$ from here I cannot move forward.

 

[anemone]

Hello everyone. I need help on this one

Prove that $ \left((a-b)^2+(b-c)^2+(c-a)^2\right)^2=2\left((a-b)^4+(b-c)^4+(c-a)^4\right)$I noticed that the leftside of the eqn when expanded would be like
$(X^2+Y^2+Z^2+2XY+2XZ+2YZ)$ from here I cannot move forward.

Hello!

This is a very good exercise problem and the very first thing you have to observe, before you can proceed would be the following:

$a-b+b-c+c-a=0$

So, if we let $x=a-b,\,y=b-c,\,z=c-a$, this implies $x+y+z=0$, which further gives us $x+y=-z$.

With this hint, I suggest you to try it because I can see how this problem can positively improve your basic mathematical skill!

 

[NotaMathPerson]

Hello!

With this hint, I suggest you to try it because I can see how this problem can positively improve your basic mathematical skill!

I agree! I was so delighted when I stumbled upon this kind of problem from a very old book. There are lots of interesting and challenging problems in that book.

 

[anemone]

I agree! I was so delighted when I stumbled upon this kind of problem from a very old book. There are lots of interesting and challenging problems in that book.

I am happy to hear that you have found so many challenges at that good book!

Hmm...so, have you been able to proceed?

 

[NotaMathPerson]

Hello this is my solution!
$x+y+z=0$
$(x^2+y^2+z^2)^2 =0$ -----> 1

Expanding 1

$\left(x^4+y^4+z^4+2x^2y^2+2x^2z^2+2y^2z^2\right)=0 $

$x^4+y^4+z^4=0 $

So,

$2x^2y^2+2x^2z^2+2y^2z^2=0 $

We know that $x+y=-z$

So $z^2=-x^2-y^2$


$2x^2y^2+2x^2(-x^2-y^2)+2y^2(-x^2-y^2)=0$

Then,

$2x^2y^2-2x^4-2x^2y^2-2y^4-2x^2y^2=0$
$-2x^2y^2-2x^4-2y^4=0$

$-2x^2y^2=2x^4+2y^4$

$x^2y^2=z^4$

$-2z^4-2x^4-2y^4=0$------> 2

Multiplying 2 by -1

$2z^4+2x^4+2y^4= 0$

Now I have

$2\left((a-b)^4+(b-c)^4+(c-a)^4\right)$

 

[anemone]

Hello this is my solution!
$x+y+z=0$
$(x^2+y^2+z^2)^2 =0$ -----> 1

Hi again, :)

I don't understand why $(x^2+y^2+z^2)^2 =0$ is true, please revise on how to expand a square of trinomial correctly because

$(x+y+z)^2\ne (x^2+y^2+z^2)^2$

and

$(x^2+y^2+z^2)^2\ne x^4+y^4+z^4+2xy+2xz+2yz$

but

$(x^2+y^2+z^2)^2=x^4+y^4+z^4+2x^2y^2+2x^2z^2+2y^2z^2$

 

[NotaMathPerson]

Hi again, :)

I don't understand why $(x^2+y^2+z^2)^2 =0$ is true, please revise on how to expand a square of trinomial correctly because

$(x+y+z)^2\ne (x^2+y^2+z^2)^2$

and

$(x^2+y^2+z^2)^2\ne x^4+y^4+z^4+2xy+2xz+2yz$

but

$(x^2+y^2+z^2)^2=x^4+y^4+z^4+2x^2y^2+2x^2z^2+2y^2z^2$

Hello
Since
$x+y+z =0$

The sum of their squares would also be equal to zero hence $x^2+y^2+z^2=0$
And the square of the sum of their squares is also zero therefore $ (x^2+y^2+z^2)^2 =0$

I forgot to square the xs ys and zs in expanding :)

 

[MarkFL]

Hello
Since
$x+y+z =0$

The sum of their squares would also be equal to zero hence $x^2+y^2+z^2=0$
And the square of the sum of their squares is also zero therefore $ (x^2+y^2+z^2)^2 =0$

I forgot to square the xs ys and zs in expanding :)

Suppose:

$$(x,y,z)=(1,2,-3)$$

Thus, we know:

$$x+y+z=0$$

Does this actually then imply:

$$x^2+y^2+z^2=0$$?

 

[NotaMathPerson]

Suppose:

$$(x,y,z)=(1,2,-3)$$

Thus, we know:

$$x+y+z=0$$

Does this actually then imply:

$$x^2+y^2+z^2=0$$?

Oh! I am missing something for sure. But I just used anemone's hint. Now I am confused again.

 

[anemone]

Hello everyone. I need help on this one

Prove that $ \left((a-b)^2+(b-c)^2+(c-a)^2\right)^2=2\left((a-b)^4+(b-c)^4+(c-a)^4\right)$I noticed that the leftside of the eqn when expanded would be like
$(X^2+Y^2+Z^2+2XY+2XZ+2YZ)$ from here I cannot move forward.

If we let $x=a-b,\,y=b-c,\,z=c-a$

then what we want to prove is: $\left(x^2+y^2+z^2\right)^2=2\left(x^4+y^4+z^4\right)$.

But we know

$x+y=-z$

$(x+y)^2=(-z)^2$

$x^2+2xy+y^2=z^2$

$x^2+y^2-z^2=-2xy$

$x^2+y^2+x^2+y^2-z^2=x^2+y^2-2xy$

$2x^2+2y^2-z^2=(x-y)^2$

$(x+y)^2(2x^2+2y^2-z^2)=(x+y)^2(x-y)^2$

$(-z)^2(2x^2+2y^2-z^2)=((x+y)(x-y))^2$

$z^2(2x^2+2y^2-z^2)=(x^2-y^2)^2$

$\therefore x^4+y^4+z^4=2x^2y^2+2x^2z^2+2y^2z^2$

Now, what can you make use of this identity?

 

[NotaMathPerson]

If we let $x=a-b,\,y=b-c,\,z=c-a$

then what we want to prove is: $\left(x^2+y^2+z^2\right)^2=2\left(x^4+y^4+z^4\right)$.

But we know

$x+y=-z$

$(x+y)^2=(-z)^2$

$x^2+2xy+y^2=z^2$

$x^2+y^2-z^2=-2xy$

$x^2+y^2+x^2+y^2-z^2=x^2+y^2-2xy$

$2x^2+2y^2-z^2=(x-y)^2$

$(x+y)^2(2x^2+2y^2-z^2)=(x+y)^2(x-y)^2$

$(-z)^2(2x^2+2y^2-z^2)=((x+y)(x-y))^2$

$z^2(2x^2+2y^2-z^2)=(x^2-y^2)^2$

$\therefore x^4+y^4+z^4=2x^2y^2+2x^2z^2+2y^2z^2$

Now, what can you make use of this identity?

Since $\left(x^2+y^2+z^2\right)^2=(x^4+y^4+z^4+2x^2y^2+2x^2z^2+2y^2z^2) $

And

$x^4+y^4+z^4=2x^2y^2+2x^2z^2+2y^2z^2$

Then

$\left(x^2+y^2+z^2\right)^2=(x^4+y^4+z^4+x^4+y^4+z^4)=2x^4+2y^4+2z^4 $

$\therefore \left(x^2+y^2+z^2\right)^2=2\left(x^4+y^4+z^4\right)$.

 

[anemone]

Since $\left(x^2+y^2+z^2\right)^2=(x^4+y^4+z^4+2x^2y^2+2x^2z^2+2y^2z^2) $

And


$\therefore x^4+y^4+z^4=2x^2y^2+2x^2z^2+2y^2z^2$

Then Since $\left(x^2+y^2+z^2\right)^2=(x^4+y^4+z^4+x^4+y^4+z^4)=2x^4+2y^4+2z^4 $


$\therefore \left(x^2+y^2+z^2\right)^2=2\left(x^4+y^4+z^4\right)$.

Correct!(Clapping) By the way, do you have anything you couldn't follow in my previous reply? If so, please write back, I would be willing to explain it to you. :D

 

[NotaMathPerson]

Correct!(Clapping) By the way, do you have anything you couldn't follow in my previous reply? If so, please write back, I would be willing to explain it to you. :D

Hello! Thanks for your help, I really appreciate it. I guess the reason I did not get it the first time around is because of lack of experience on solving problems like this one. I really should practice more, and I am hoping MHB people would help again when I get stuck. Thanks again for the assistance!