If `(loga)/(y-z)=(logb)/(z-x)=(logc)/(x-y)` and `a^(y^(2)+yz+z^(2))b^(z^(2)+zx+x^(2))c^(x^(2)+xy+y^(2))=` ….

Answer

To solve the given problem, we start with the equations provided:

Given:
\[ \frac{\log a}{y - z} = \frac{\log b}{z - x} = \frac{\log c}{x - y} = t \]

From this, we can express \(\log a\), \(\log b\), and \(\log c\) in terms of \(t\):
1. \(\log a = t(y - z)\)
2. \(\log b = t(z - x)\)
3. \(\log c = t(x - y)\)

Next, we need to evaluate the expression:
\[ a^{y^2 + yz + z^2} b^{z^2 + zx + x^2} c^{x^2 + xy + y^2} \]

Using the properties of logarithms, we can rewrite the expression:
\[ \log\left(a^{y^2 + yz + z^2} b^{z^2 + zx + x^2} c^{x^2 + xy + y^2}\right) = (y^2 + yz + z^2) \log a + (z^2 + zx + x^2) \log b + (x^2 + xy + y^2) \log c \]

Substituting the values of \(\log a\), \(\log b\), and \(\log c\):
\[ = (y^2 + yz + z^2)(t(y - z)) + (z^2 + zx + x^2)(t(z - x)) + (x^2 + xy + y^2)(t(x - y)) \]

Factoring out \(t\):
\[ = t \left[(y^2 + yz + z^2)(y - z) + (z^2 + zx + x^2)(z - x) + (x^2 + xy + y^2)(x - y)\right] \]

Now, we analyze the expression inside the brackets. Notice that it is a cyclic sum of the form:
\[ S = (y^2 + yz + z^2)(y - z) + (z^2 + zx + x^2)(z - x) + (x^2 + xy + y^2)(x - y) \]

This sum \(S\) can be shown to equal zero because it is a symmetric polynomial in \(x\), \(y\), and \(z\) that cancels out when summed cyclically.

Thus, we have:
\[ t \cdot S = 0 \implies \log\left(a^{y^2 + yz + z^2} b^{z^2 + zx + x^2} c^{x^2 + xy + y^2}\right) = 0 \]

This implies:
\[ a^{y^2 + yz + z^2} b^{z^2 + zx + x^2} c^{x^2 + xy + y^2} = 1 \]

Finally, we conclude that:
\[ \boxed{1} \]