You need to use mathematical induction, hence, you need to start with the first step of the method, the basis, such that:

`{(x + y + z = 1),(x^2 + y^2 + z^2 = 2),(x^3 + y^3 + z^3 = 3):}`

Taking the next step requires you to prove that if the statement holds for each n, hence, it also holds for `n + 1` , such that:

`x^n + y^n + z^n = n => x^(n+1) + y^(n+1) + z^(n+1) = n+1`

You need to multiplicate by x the valid statement `x^n + y^n + z^n = n` , such that:

`x^(n+1)+ x*y^n + x*z^n = x*n`

You need to multiplicate by y the valid statement `x^n + y^n + z^n = n` , such that:

`y*x^n + y^(n+1) + y*z^n = y*n`

You need to multiplicate by z the valid statement `x^n + y^n + z^n = n` , such that:

`z*x^n + z*y^n + z^(n+1) = z*n`

Adding the relations yields:

`x^(n+1) + y^(n+1) + z^(n+1) + x^n(y + z) + y^n(x + z) + z^n(x + y) = n*(x + y + z)`

Since `x + y + z = 1` yields:

`x^(n+1) + y^(n+1) + z^(n+1) + x^n(y + z) + y^n(x + z) + z^n(x + y) = n`

You need to use the exponential laws such that:

`x^n*x + y^n*y + z^n*z + x^n(y + z) + y^n(x + z) + z^n(x + y) = n`

Factoring out `x^n, y^n, z^n` , yields:

`x^n(x + y + z) + y^n(x + y + z) + z^n(x + y + z) = n`

Since `x + y + z = 1` yields:

`x^n + y^n + z^n = n`

**Hence, testing if `x^(n+1) + y^(n+1) + z^(n+1) = n+1 ` holds yields the valid statement `x^n + y^n + z^n = n,` hence, reasoning by analogy yields `x^4 + y^4 + z^4 = 4.` **

x^(n+1) + y^(n+1) + z^(n+1) = n+1

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