#### lordofdrinks

x+y+z=1
x^2+y^2+z^2=2
x^3+y^3+z^3=3
x^4+y^4+z^4=?

answer is not 4 thanks.

#### romsek

MHF Helper
$x^4+y^4+z^4=\dfrac{25}{6}$

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#### lordofdrinks

Thanks for your answer but I don't understand this. Could you describe in more detail? Thanks.

#### DenisB

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#### DenisB

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#### DenisB

x+y+z=1
x^2+y^2+z^2=2
1st equation doubled: 2(x + y + z) = 2
SO:
x^2 + y^2 + z^2 = 2x + 2y + 2z
Rearranged:
x^2 - 2x + y^2 + z^2 - 2y - 2z = 0
Let k = y^2 + z^2 - 2y - 2z; then:

x^2 - 2x + k = 0
Solve for x:
x = 1 +- SQRT(1 - k)

Poor x has very little chance of being real!!

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#### Idea

express in terms of elementary symmetric functions. Let

$$\displaystyle x+y+z=s$$

$$\displaystyle x y + y z + z x = r$$

$$\displaystyle x y z=p$$

we get

$$\displaystyle x^2+y^2+z^2=s^2-2r=2$$

$$\displaystyle x^3+y^3+z^3=3p+s^3-3s r=3$$

since $$\displaystyle s=1$$ the above equations give $$\displaystyle r=-1/2, p= 1/6$$

therefore

$$\displaystyle x^4+y^4+z^4=4p s+s^4-4s^2r+2r^2=\frac{25}{6}$$

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