Roselyn Daniel

Answered

2022-07-16

Fraction Decomposition

I have the following problem:

Suppose $x+y+z=0$. Show that

$\frac{{x}^{5}+{y}^{5}+{z}^{5}}{5}=\frac{{x}^{3}+{y}^{3}+{z}^{3}}{3}\times \frac{{x}^{2}+{y}^{2}+{z}^{2}}{2}$

and

$\frac{{x}^{7}+{y}^{7}+{z}^{7}}{7}=\frac{{x}^{2}+{y}^{2}+{z}^{2}}{2}\times \frac{{x}^{5}+{y}^{5}+{z}^{5}}{5}$

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?

I have the following problem:

Suppose $x+y+z=0$. Show that

$\frac{{x}^{5}+{y}^{5}+{z}^{5}}{5}=\frac{{x}^{3}+{y}^{3}+{z}^{3}}{3}\times \frac{{x}^{2}+{y}^{2}+{z}^{2}}{2}$

and

$\frac{{x}^{7}+{y}^{7}+{z}^{7}}{7}=\frac{{x}^{2}+{y}^{2}+{z}^{2}}{2}\times \frac{{x}^{5}+{y}^{5}+{z}^{5}}{5}$

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

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:

$(x+y+z{)}^{3}={x}^{3}+{y}^{3}+{z}^{3}+3{x}^{2}y+3{x}^{2}z+3{y}^{2}x+3{y}^{2}z+3{z}^{2}x+3{z}^{2}y+6xyz$

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

$0={x}^{3}+{y}^{3}+{z}^{3}+3{x}^{2}(-x)+3{y}^{2}(-y)+3{z}^{2}(-z)+6xyz$

and even better than that!

$\begin{array}{rl}0& =-2({x}^{3}+{y}^{3}+{z}^{3})+6xyz\\ xyz& =\frac{{x}^{3}+{y}^{3}+{z}^{3}}{3}.\end{array}$

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 ${S}_{111}=xyz$ and ${S}_{k}={x}^{k}+{y}^{k}+{z}^{k}$, among which you have a first relation here, you'll need other symmetric forms like ${S}_{11}=xy+xz+yz$

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

$\begin{array}{rl}{S}_{2}& =-2{S}_{11}\\ {S}_{3}& =3{S}_{111}\\ {S}_{4}& =\frac{1}{2}{S}_{2}^{2}\end{array}$

and also some useful observations like

$\begin{array}{rl}{S}_{kl}& ={S}_{k}{S}_{l}-{S}_{k+l}\\ {S}_{(a+c)(b+c)c}& ={S}_{ccc}{S}_{ab}={S}_{111}^{c}{S}_{ab}\end{array}$

With these you're one little theorem away from ${S}_{5}/5={S}_{3}{S}_{2}/6$ and ${S}_{7}/7={S}_{5}{S}_{2}/10$

$(x+y+z{)}^{3}={x}^{3}+{y}^{3}+{z}^{3}+3{x}^{2}y+3{x}^{2}z+3{y}^{2}x+3{y}^{2}z+3{z}^{2}x+3{z}^{2}y+6xyz$

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

$0={x}^{3}+{y}^{3}+{z}^{3}+3{x}^{2}(-x)+3{y}^{2}(-y)+3{z}^{2}(-z)+6xyz$

and even better than that!

$\begin{array}{rl}0& =-2({x}^{3}+{y}^{3}+{z}^{3})+6xyz\\ xyz& =\frac{{x}^{3}+{y}^{3}+{z}^{3}}{3}.\end{array}$

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 ${S}_{111}=xyz$ and ${S}_{k}={x}^{k}+{y}^{k}+{z}^{k}$, among which you have a first relation here, you'll need other symmetric forms like ${S}_{11}=xy+xz+yz$

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

$\begin{array}{rl}{S}_{2}& =-2{S}_{11}\\ {S}_{3}& =3{S}_{111}\\ {S}_{4}& =\frac{1}{2}{S}_{2}^{2}\end{array}$

and also some useful observations like

$\begin{array}{rl}{S}_{kl}& ={S}_{k}{S}_{l}-{S}_{k+l}\\ {S}_{(a+c)(b+c)c}& ={S}_{ccc}{S}_{ab}={S}_{111}^{c}{S}_{ab}\end{array}$

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

Expert

2022-07-18Added 5 answers

The first identity does not involve much algebra, if you write $z=-(x+y)$

${z}^{5}+({x}^{5}+{y}^{5})=-5{x}^{4}y-10{x}^{3}{y}^{2}-10{x}^{2}{y}^{3}-5x{y}^{4}$

where the $(-{x}^{5}-{y}^{5})$ cancels out in the binomial exapnsion. A similar thing happends in the cubes term.

So the identity reads

$-{x}^{4}y-2{x}^{3}{y}^{2}-2{x}^{2}{y}^{3}-x{y}^{4}=(-{x}^{2}y-x{y}^{2})\frac{({x}^{2}+{y}^{2}+{x}^{2}+2xy+{y}^{2})}{2}$

and of the six terms on the right, two pairs add together, and you get the left side.

${z}^{5}+({x}^{5}+{y}^{5})=-5{x}^{4}y-10{x}^{3}{y}^{2}-10{x}^{2}{y}^{3}-5x{y}^{4}$

where the $(-{x}^{5}-{y}^{5})$ cancels out in the binomial exapnsion. A similar thing happends in the cubes term.

So the identity reads

$-{x}^{4}y-2{x}^{3}{y}^{2}-2{x}^{2}{y}^{3}-x{y}^{4}=(-{x}^{2}y-x{y}^{2})\frac{({x}^{2}+{y}^{2}+{x}^{2}+2xy+{y}^{2})}{2}$

and of the six terms on the right, two pairs add together, and you get the left side.

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