1. **State the problem:** We need to prove that $$\frac{1}{p} + \frac{1}{q} + \frac{1}{x} = \frac{1}{x+p+q}$$. 2. **Analyze the equation:** The left side is the sum of three fractions with denominators $p$, $q$, and $x$. The right side is a single fraction with denominator $x+p+q$. 3. **Find a common denominator on the left side:** The common denominator is $pqx$. 4. **Rewrite the left side with the common denominator:** $$\frac{1}{p} + \frac{1}{q} + \frac{1}{x} = \frac{qx}{pqx} + \frac{px}{pqx} + \frac{pq}{pqx} = \frac{qx + px + pq}{pqx}$$ 5. **Simplify the numerator:** $$qx + px + pq = xq + xp + pq$$ 6. **Compare the left side and right side:** The left side is $$\frac{xq + xp + pq}{pqx}$$ and the right side is $$\frac{1}{x+p+q}$$. 7. **Check if the two sides are equal:** Cross-multiply to verify: $$ (xq + xp + pq)(x + p + q) \stackrel{?}{=} pqx $$ 8. **Expand the left side:** $$ (xq)(x) + (xq)(p) + (xq)(q) + (xp)(x) + (xp)(p) + (xp)(q) + (pq)(x) + (pq)(p) + (pq)(q) $$ $$ = x^2 q + x p q + x q^2 + x^2 p + x p^2 + x p q + p q x + p^2 q + p q^2 $$ 9. **Simplify terms:** Notice terms like $x p q$ appear multiple times. 10. **Conclusion:** The expanded left side is a polynomial of degree 2, while the right side is $pqx$, degree 3 in denominator. They are not equal in general. **Therefore, the equation $$\frac{1}{p} + \frac{1}{q} + \frac{1}{x} = \frac{1}{x+p+q}$$ is not true for general values of $p$, $q$, and $x$.** If you meant a different relation or need help with a specific problem, please clarify.