Prove (a^5 + b^5 + c^5)/5 = (a^3 + b^3 + c^3)/3 * (a^2 + b^2 + c^2)/2

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In summary, the equation $x^3+\beta x-\gamma=0$ can be solved for $a,b,c$ roots using the Sum-of-Squares Method. When solving for $a^5+b^5+c^5$, it is necessary to restrict the variables to be a,b,c=0.
  • #1
mente oscura
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Hello.

Prove:

[tex]\dfrac{a^5+b^5+c^5}{5}=\dfrac{a^3+b^3+c^3}{3} \ \dfrac{a^2+b^2+c^2}{2}[/tex]

Regards.
 
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  • #2
Re: prove (a^5+b^5+c^5)/5=(a^3+b^3+c^3)/3 * (a^2+b^2+c^2)/2

I believe we require the restriction that $a+b+c=0$. Note that for $a=b=c=1$ the equation does not hold.

Here is my solution, which I have slightly modified by changing the variables, from my solution to essentially the same problem posted by anemone here:

http://mathhelpboards.com/challenge-questions-puzzles-28/prove-s_5-5%3D-s_3-3s_2-2-a-6692.html

Let $S_n=a^n+b^n+c^n$, where $S_1=0$.

If we view $S_n$ as a recursive algorithm, we see that it must come from the characteristic equation:

\(\displaystyle (r-a)(r-b)(r-c)=0\)

\(\displaystyle r^3-(a+b+c)r^2+(ab+ac+bc)r-abc=0\)

Since $a+b+c=S_1=0$, we obtain the following recursion:

\(\displaystyle S_{n+3}=-(ab+ac+bc)S_{n+1}+abcS_{n}\)

Now, observing we may write:

\(\displaystyle (a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)\)

\(\displaystyle 0=S_2+2(ab+ac+bc)\)

\(\displaystyle -(ab+ac+bc)=\frac{S_2}{2}\)

Also, we find:

\(\displaystyle (a+b+c)^3=-2\left(a^3+b^3+c^3 \right)+3\left(a^2+b^2+c^2 \right)(a+b+c)+6abc\)

\(\displaystyle 0=-2S_3+6abc\)

\(\displaystyle abc=\frac{S^3}{3}\)

And so our recursion may be written:

\(\displaystyle S_{n+3}=\frac{S_2}{2}S_{n+1}+\frac{S_3}{3}S_{n}\)

Letting $n=2$, we then find:

\(\displaystyle S_{5}=\frac{S_2}{2}S_{3}+\frac{S_3}{3}S_{2}\)

\(\displaystyle S_{5}=\frac{5}{6}S_2S_{3}\)

\(\displaystyle \frac{S_5}{5}=\frac{S_3}{3}\cdot\frac{S_2}{2}\)

Shown as desired.
 
  • #3
Re: prove (a^5+b^5+c^5)/5=(a^3+b^3+c^3)/3 * (a^2+b^2+c^2)/2

mente oscura said:
Hello.

Prove:

[tex]\dfrac{a^5+b^5+c^5}{5}=\dfrac{a^3+b^3+c^3}{3} \ \dfrac{a^2+b^2+c^2}{2}[/tex]

Regards.
Thanks to I Like Serena to point out a mistake in my original post.
CORRECTED POST:
As Mark mentioned, we require $a+b+c=0$ to prove the above.

Let $a,b$ and $c$ be roots of $x^3+\beta x-\gamma=0$.

I will use $\sum a$ to denote $a+b+c$, $\sum a^2$ to denote $a^2+b^2+c^2$, and so on.

Further, let's use $\sum ab$ to denote $ab+bc+ca$.

Then $\sum a^2=(\sum a)^2-2(\sum ab)=-2\beta$.

Thus $$\sum a^2=-2\beta$$

Since $x^3+\beta x-\gamma=0$ is satisfied by $a,b,c$, we get $\sum a^3+\beta(\sum a)-3\gamma=0$, giving $$\sum a^3=3\gamma$$.

Also $x^2(x^3+\beta x-\gamma)=0$ is satisfied by $a,b,c$.

So $x^5+\beta x^3-\gamma x^2=0$ is satisfied by $a,b,c$.

Thus $\sum a^5+\beta(\sum a^3)-\gamma(\sum a^2)=0$.

This leads to $$\sum a^5=-3\beta\gamma-2\beta\gamma=-5\beta\gamma$$.

From here the desired equality easily follows.

ORIGINAL POST:
As Mark mentioned, we require $a+b+c=0$ to prove the above.

Let $a,b$ and $c$ be roots of $x^3+bx-c=0$.

I will use $\sum a$ to denote $a+b+c$, $\sum a^2$ to denote $a^2+b^2+c^2$, and so on.

Further, let's use $\sum ab$ to denote $ab+bc+ca$.

Then $\sum a^2=(\sum a)^2-2(\sum ab)=-2b$.

Thus $$\sum a^2=-2b$$

Since $x^3+bx-c=0$ is satisfied by $a,b,c$, we get $\sum a^3+b(\sum a)-3c=0$, giving $$\sum a^3=3c$$.

Also $x^2(x^3+bx-c)=0$ is satisfied by $a,b,c$.

So $x^5+bx^3-cx^2=0$ is satisfied by $a,b,c$.

Thus $\sum a^5+b(\sum a^3)-c(\sum a^2)=0$.

This leads to $$\sum a^5=-3bc-2bc=-5bc$$.

From here the desired equality easily follows.
 
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  • #4
Re: prove (a^5+b^5+c^5)/5=(a^3+b^3+c^3)/3 * (a^2+b^2+c^2)/2

caffeinemachine said:
As Mark mentioned, we require $a+b+c=0$ to prove the above.

Let $a,b$ and $c$ be roots of $x^3+bx-c=0$.

I will use $\sum a$ to denote $a+b+c$, $\sum a^2$ to denote $a^2+b^2+c^2$, and so on.

Further, let's use $\sum ab$ to denote $ab+bc+ca$.

Then $\sum a^2=(\sum a)^2-2(\sum ab)=-2b$.

Thus $$\sum a^2=-2b$$

Since $x^3+bx-c=0$ is satisfied by $a,b,c$, we get $\sum a^3+b(\sum a)-3c=0$, giving $$\sum a^3=3c$$.

Also $x^2(x^3+bx-c)=0$ is satisfied by $a,b,c$.

So $x^5+bx^3-cx^2=0$ is satisfied by $a,b,c$.

Thus $\sum a^5+b(\sum a^3)-c(\sum a^2)=0$.

This leads to $$\sum a^5=-3bc-2bc=-5bc$$.

From here the desired equality easily follows.

Nice and elegant. :cool:
However, shouldn't your equation be $x^3+(\sum ab)x-abc=0$?
With this modification, the same argument works out.
 
  • #5
Re: prove (a^5+b^5+c^5)/5=(a^3+b^3+c^3)/3 * (a^2+b^2+c^2)/2

I like Serena said:
Nice and elegant. :cool:
However, shouldn't your equation be $x^3+(\sum ab)x-abc=0$?
With this modification, the same argument works out.
Ah! What I meant was let $x^3+\beta x-\gamma =0$ be the cubic whose roots are $a,b$ and $c$. There was a glaring clash in notation which I didn't see.
 
Last edited:
  • #6
Re: prove (a^5+b^5+c^5)/5=(a^3+b^3+c^3)/3 * (a^2+b^2+c^2)/2

Clever use of powersum identities, caffeinemachine!
 
  • #7
Re: prove (a^5+b^5+c^5)/5=(a^3+b^3+c^3)/3 * (a^2+b^2+c^2)/2

Inspired by caffeinemachine, I decided to explore this a bit more.
I've looked up Newton's Identities, that define:
\begin{array}{}
p_k &=& \sum a^k \\
e_k &=& \text{sum of all distinct products of k distinct variables}
\end{array}

after which we have the identities:
\begin{array}{}
p_1 &=& e_1 \\
p_2 &=& e_1p_1 - 2e_2 \\
p_3 &=& e_1p_2 - e_2p_1 + 3e_3 \\
p_5 &=& e_1p_4 - e_2p_3 + e_3p_2 - e_4p_1 + 5e_5
\end{array}

Following caffeinemachine's notation, this becomes (with $Σ a = 0$):
\begin{array}{}
Σ a &=& Σ a
&=& 0 \\
\textstyle Σ a^2 &=& Σ a Σ a - 2Σ ab
&=& -2 Σ ab \\
Σ a^3 &=& Σ a Σ a^2 - Σ ab Σ a + 3abc
&=& 3abc \\
Σ a^5 &=& Σ a Σ a^4 - Σ ab Σ a^3 + abc Σ a^2 - 0 + 0 \\
&=& -Σ ab \cdot 3abc + abc \cdot -2 Σ ab \\
&=& -5abc Σ ab
\end{array}

The equality to prove follows immediately.
 
  • #8
Re: prove (a^5+b^5+c^5)/5=(a^3+b^3+c^3)/3 * (a^2+b^2+c^2)/2

mathbalarka said:
Clever use of powersum identities, caffeinemachine!
Thanks. :)
 
  • #9
Re: prove (a^5+b^5+c^5)/5=(a^3+b^3+c^3)/3 * (a^2+b^2+c^2)/2

Hello.

I'am omitted the restriction that [tex]a, \ b, \ c \in{Z} \ and \ a+b+c=0[/tex]. I'm sorry.:eek:

This problem, to be propose in another forum of mathematics in the Spanish language, and I resolved it in the following way: A little to the brute force.:eek:

[tex](a+b+c)^5=a^5+b^5+c^5+5(a+b)(a+c)(b+c)(a^2+b^2+c^2+ab+ac+bc)[/tex]

[tex](a+b+c)^3=a^3+b^3+c^3+3(a+b)(a+c)(b+c)[/tex]

[tex](a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc[/tex][tex]\dfrac{a^5+b^5+c^5}{5}=-(a+b)(a+c)(b+c)(a^2+b^2+c^2+ab+ac+bc)[/tex]

[tex]\dfrac{a^5+b^5+c^5}{5}=-[-\dfrac{a^3+b^3+c^3}{3} \ (a^2+b^2+c^2+ab+ac+bc)][/tex]

[tex]\dfrac{a^5+b^5+c^5}{5}=-[-\dfrac{a^3+b^3+c^3}{3}(a^2+b^2+c^2-\dfrac{a^2+b^2+c^2}{2})][/tex]

[tex]\dfrac{a^5+b^5+c^5}{5}=\dfrac{a^3+b^3+c^3}{3} \ \dfrac{a^2+b^2+c^2}{2}[/tex]

I expose hidden, in case someone wants to keep trying the prove.

Regards.
 

FAQ: Prove (a^5 + b^5 + c^5)/5 = (a^3 + b^3 + c^3)/3 * (a^2 + b^2 + c^2)/2

What is the equation to prove?

The equation to prove is (a^5 + b^5 + c^5)/5 = (a^3 + b^3 + c^3)/3 * (a^2 + b^2 + c^2)/2

What are the variables a, b, and c?

The variables a, b, and c represent any real numbers.

What are the steps to prove this equation?

The steps to prove this equation involve using algebraic manipulation and the properties of exponents to simplify and rearrange the terms until they are equal on both sides.

Can this equation be proven for any values of a, b, and c?

Yes, this equation can be proven for any real numbers a, b, and c as long as they are not equal to 0.

Why is this equation important?

This equation is important because it shows a relationship between the sums of powers of numbers, and it can be used in various mathematical and scientific applications.

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