Answer: To find the value of x^3 + y^3 + z^3 - 3xyz, we can use the identity known as the "sum of cubes" formula, which states:
a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc).
In this case, a = x, b = y, and c = z. We are given that x + y + z = 1, so we can substitute this into the formula:
x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - xz - yz).
We are also given that x^2 + y^2 + z^2 = 83, so we substitute this value as well:
x^3 + y^3 + z^3 - 3xyz = (1)(83 - xy - xz - yz).
Now, we need to find the values of xy, xz, and yz. To do this, we can square the equation x + y + z = 1:
(x + y + z)^2 = 1^2
x^2 + y^2 + z^2 + 2(xy + xz + yz) = 1.
Since we know that x^2 + y^2 + z^2 = 83, we can substitute this into the equation and solve for xy + xz + yz:
83 + 2(xy + xz + yz) = 1
2(xy + xz + yz) = 1 - 83
2(xy + xz + yz) = -82
xy + xz + yz = -41.
Now, substitute this value back into the expression we found earlier:
x^3 + y^3 + z^3 - 3xyz = (1)(83 - (-41))
x^3 + y^3 + z^3 - 3xyz = 124.
Therefore, the value of x^3 + y^3 + z^3 - 3xyz is 124.