Decide what values of the variable cannot possibly be solutions for the equation. Do not solve.
\[
\frac{1}{x-1}+\frac{1}{x+2}=\frac{1}{x^{2}+x-2}
\]
What values of $x$ cannot be solutions of the equation?

Step 1 ：Consider the equation \(\frac{1}{x-1}+\frac{1}{x+2}=\frac{1}{x^{2}+x-2}\).

Step 2 ：The values of \(x\) that cannot be solutions of the equation are the ones that make the denominator of any of the fractions equal to zero, because division by zero is undefined in mathematics.

Step 3 ：Therefore, we need to find the values of \(x\) that make \(x-1=0\), \(x+2=0\), or \(x^{2}+x-2=0\).

Step 4 ：Solving these equations, we find that the values of \(x\) that make the denominators of the fractions equal to zero are \(1\) and \(-2\).

Step 5 ：Final Answer: The values of \(x\) that cannot be solutions of the equation are \(\boxed{1}\) and \(\boxed{-2}\).