- #1

- 328

- 0

x+y+z=0

x

^{3}+y

^{3}+z

^{3}=3

x

^{5}+y

^{5}+z

^{5}=15

x

^{2}+y

^{2}+z

^{2}=?

An answer must be supported with justification (so that I know that you didn't guess and get lucky).

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter ƒ(x)
- Start date

In summary, the problem presents three equations with three unknown variables, x, y, and z. The first equation is x+y+z=0, the second equation is x3+y3+z3=3, and the third equation is x5+y5+z5=15. The question asks for the value of x2+y2+z2. This problem can be easily solved using a website like Wolfram Alpha or by solving the simultaneous equations manually. The answer to x2+y2+z2 is a tricky one, but can be easily obtained once the values of x, y, and z are known.

- #1

- 328

- 0

x+y+z=0

x

x

x

An answer must be supported with justification (so that I know that you didn't guess and get lucky).

Physics news on Phys.org

- #2

- 45

- 0

I'm going to ask my math teacher :)

- #3

- 101

- 0

ƒ(x) said:

x+y+z=0

x^{3}+y^{3}+z^{3}=3

x^{5}+y^{5}+z^{5}=15

x^{2}+y^{2}+z^{2}=?

An answer must be supported with justification (so that I know that you didn't guess and get lucky).

Unless you can demonstrate that there is a clever shortcut to this problem which would raise it to the level of a brainteaser, this is just a set of simultaneous equations which is easily solvable with, for example, the Wolfram Alpha website.

- #4

- 328

- 0

regor60 said:Unless you can demonstrate that there is a clever shortcut to this problem which would raise it to the level of a brainteaser, this is just a set of simultaneous equations which is easily solvable with, for example, the Wolfram Alpha website.

I had no idea that such a site even existed.

- #5

- 45

- 0

How do they solve it ?

- #6

- 31

- 0

I got the answer but i cant´t get the values os x, y and z, but i got that x.y.z =

x.y.z=1

- #7

- 31

- 0

x= 2 cos (20)

y= 2 cos (140)

z= 2 cos (260)

the answer to the question is 6

y= 2 cos (140)

z= 2 cos (260)

the answer to the question is 6

- #8

- 31

- 0

At a glance:

x = -y-z

x<2 and positive

y not equal z

y and z are negative

x.y.z = 1

x<2 and positive

y not equal z

y and z are negative

x.y.z = 1

- #9

- 31

- 0

^2 = 14 --> ?

^3 = 18 --> 3

^5 = 210 --> 15

so 14 . 18 . 15 = ? . 3 . 210

? = 6

That´s the tricky way...

Share:

- Replies
- 14

- Views
- 1K

- Replies
- 1

- Views
- 602

- Replies
- 3

- Views
- 537

- Replies
- 16

- Views
- 1K

- Replies
- 5

- Views
- 614

- Replies
- 372

- Views
- 13K

- Replies
- 87

- Views
- 4K

- Replies
- 11

- Views
- 677

- Replies
- 3

- Views
- 2K

- Replies
- 1

- Views
- 548