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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.
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.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.
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.
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.I like Serena said:Nice and elegant.
However, shouldn't your equation be $x^3+(\sum ab)x-abc=0$?
With this modification, the same argument works out.
Thanks. :)mathbalarka said:Clever use of powersum identities, caffeinemachine!
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
The variables a, b, and c represent any real numbers.
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.
Yes, this equation can be proven for any real numbers a, b, and c as long as they are not equal to 0.
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.