MHB Proving Algebraic Identity: How to Simplify Complex Equations

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The discussion revolves around proving the algebraic identity \( \left((a-b)^2+(b-c)^2+(c-a)^2\right)^2=2\left((a-b)^4+(b-c)^4+(c-a)^4\right) \). Participants suggest using the substitutions \( x = a-b \), \( y = b-c \), and \( z = c-a \), noting that \( x+y+z=0 \). Several users provide insights on expanding the squares and simplifying the expressions, leading to the conclusion that \( (x^2+y^2+z^2)^2=2(x^4+y^4+z^4) \). The discussion emphasizes the importance of practice in tackling such algebraic problems, with users expressing appreciation for the collaborative help. Overall, the thread highlights the value of peer support in understanding complex mathematical concepts.
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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.
 
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NotaMathPerson said:
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!
 
anemone said:
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.
 
NotaMathPerson said:
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! :cool:

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

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

$(x^2+y^2+z^2)^2=x^4+y^4+z^4+2x^2y^2+2x^2z^2+2y^2z^2$
 
anemone said:
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 :)
 
NotaMathPerson said:
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:

$$x^2+y^2+z^2=0$$?
 
MarkFL said:
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.
 
  • #10
NotaMathPerson said:
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?
 
  • #11
anemone said:
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 $

$\therefore \left(x^2+y^2+z^2\right)^2=2\left(x^4+y^4+z^4\right)$.
 
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  • #12
NotaMathPerson said:
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
 
  • #13
anemone said:
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!
 
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