Fraction DecompositionI have the following problem:Suppose x + y + z = 0. Show that...

Roselyn Daniel

Roselyn Daniel

Answered

2022-07-16

Fraction Decomposition
I have the following problem:
Suppose x + y + z = 0. Show that
x 5 + y 5 + z 5 5 = x 3 + y 3 + z 3 3 × x 2 + y 2 + z 2 2
and
x 7 + y 7 + z 7 7 = x 2 + y 2 + z 2 2 × x 5 + y 5 + z 5 5
I thought this was a fun problem to tackle, but I've been stuck on this for hours, and I can't seem to get anywhere. I've tried expanding, but no matter what I do, I just get a huge mess.
I also cannot find how to incorporate the initial assumption ( x + y + z = 0) anywhere. Have I missed something?

Answer & Explanation

Jaylynn Huffman

Jaylynn Huffman

Expert

2022-07-17Added 14 answers

You've surely been trying to express one of the unknowns using the others and plugging in. Try to preserve the symmetry between them instead:
( x + y + z ) 3 = x 3 + y 3 + z 3 + 3 x 2 y + 3 x 2 z + 3 y 2 x + 3 y 2 z + 3 z 2 x + 3 z 2 y + 6 x y z
But we know that the left hand side is a zero and, moreover, wherever something like x + z appears we can replace that by a y (again, strictly keeping exchange symmetry). So the same equation can be written as
0 = x 3 + y 3 + z 3 + 3 x 2 ( x ) + 3 y 2 ( y ) + 3 z 2 ( z ) + 6 x y z
and even better than that!
0 = 2 ( x 3 + y 3 + z 3 ) + 6 x y z x y z = x 3 + y 3 + z 3 3 .
Surely replacing the right hand side by the left in your problem will make the multiplications simpler! Try to proceed from here :-)
Hint: apart from S 111 = x y z and S k = x k + y k + z k , among which you have a first relation here, you'll need other symmetric forms like S 11 = x y + x z + y z
Update: I have been able to derive both your equations this way so it's guaranteed to work. It just takes some time. There's a whole theory about monomial and power-sum symmetric polynomials but I didn't want to go into that.
Starting from S 1 = 0, you can derive things like
S 2 = 2 S 11 S 3 = 3 S 111 S 4 = 1 2 S 2 2
and also some useful observations like
S k l = S k S l S k + l S ( a + c ) ( b + c ) c = S c c c S a b = S 111 c S a b
With these you're one little theorem away from S 5 / 5 = S 3 S 2 / 6 and S 7 / 7 = S 5 S 2 / 10
Aleah Booth

Aleah Booth

Expert

2022-07-18Added 5 answers

The first identity does not involve much algebra, if you write z = ( x + y )
z 5 + ( x 5 + y 5 ) = 5 x 4 y 10 x 3 y 2 10 x 2 y 3 5 x y 4
where the ( x 5 y 5 ) cancels out in the binomial exapnsion. A similar thing happends in the cubes term.
So the identity reads
x 4 y 2 x 3 y 2 2 x 2 y 3 x y 4 = ( x 2 y x y 2 ) ( x 2 + y 2 + x 2 + 2 x y + y 2 ) 2
and of the six terms on the right, two pairs add together, and you get the left side.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?