Matematické Fórum

vanok · 27. 11. 2018 01:24

If $a+b+c=3$ and
$a^2+b^2+c^2=5$ and
$a^3+b^3+c^3=7$

what is $a^4+b^4+c^4$ ? And how is it done?

vanok · 28. 11. 2018 04:50

Hint 1
The answer is $9$ .
Prove the.

vanok · 28. 11. 2018 07:20 — Editoval vanok (29. 11. 2018 04:58)

Hint 2.
We pose $x=a; y=b; z=c$ .
You can use the elementary symmetric polynomials: $x+y+z, xy+yz+xz, xyz$ .
Note that $xy+yz+xz=\frac 12[(x+y+z)^2-(x^2+y^2+z^2)]=\frac 12 (3^2-5)=....$

Continue!

vanok · 28. 11. 2018 12:35 — Editoval vanok (29. 11. 2018 04:55)

Hint 3.
Note that
$xyz=\frac 16 [(x+y+z)^3-3(x+y+z)(x^2+y^2+z^2)+2(x^3+y^3+z^3)]$

vanok · 28. 11. 2018 18:16

Hint 4.
To express $x^4+y^4+z^4$ Use idea of Hint 2 two of times.

You see, it’s not that hard. 👍

vanok · 29. 11. 2018 05:16 — Editoval vanok (29. 11. 2018 08:43)

Note.
This method determined $x^4+y^4+z^4$ without calculating $x,y,z$

Another method make use of third-order equation. (Which may be sometimes more complicated that the other metod)

Matematické Fórum

#1 27. 11. 2018 01:24

Podla intrnetu

#2 28. 11. 2018 04:50

Re: Podla intrnetu

#3 28. 11. 2018 07:20 — Editoval vanok (29. 11. 2018 04:58)

Re: Podla intrnetu

#4 28. 11. 2018 12:35 — Editoval vanok (29. 11. 2018 04:55)

Re: Podla intrnetu

#5 28. 11. 2018 18:16

Re: Podla intrnetu

#6 29. 11. 2018 05:16 — Editoval vanok (29. 11. 2018 08:43)

Re: Podla intrnetu

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