If $\frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1$ and $\frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0$ where $a, b, c, p, q, r$ are non-zero real numbers, then $\frac{p^2}{a^2} + \frac{q^2}{b^2} + \frac{r^2}{c^2}$ is equal to

### Concept & Formula This problem is a disguised application of the algebraic identity for the square of a trinomial: $$(X + Y + Z)^2 = X^2 + Y^2 + Z^2 + 2(XY + YZ + ZX)$$ ### Step-by-Step Solution * To make the problem visually simpler, substitute the fractions with capital variables: Let $X = \frac{p}{a}$, $Y = \frac{q}{b}$, and $Z = \frac{r}{c}$. * The first given equation becomes: $$X + Y + Z = 1$$ * The second given equation represents their reciprocals: $$\frac{1}{X} + \frac{1}{Y} + \frac{1}{Z} = 0$$ * Combine the reciprocal equation into a single fraction by finding a common denominator ($XYZ$): $$\frac{YZ + XZ + XY}{XYZ} = 0$$ * Since the fraction equals zero, its numerator must be zero: $$XY + YZ + ZX = 0$$ * Now we need to find the value of $\frac{p^2}{a^2} + \frac{q^2}{b^2} + \frac{r^2}{c^2}$, which translates to finding $X^2 + Y^2 + Z^2$. * Use the trinomial square identity: $$(X + Y + Z)^2 = X^2 + Y^2 + Z^2 + 2(XY + YZ + ZX)$$ * Substitute the known values ($X+Y+Z=1$ and $XY+YZ+ZX=0$): $$1^2 = X^2 + Y^2 + Z^2 + 2(0)$$ $$1 = X^2 + Y^2 + Z^2$$ ### Exam Strategy & Shortcut By using the substitution variable method (changing complex fractions into $X, Y, Z$), you reduce visual clutter, which drastically decreases the cognitive load and prevents transcription errors under exam pressure. ### Common Pitfall A standard error is misinterpreting $\frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0$ and wrongly assuming it implies the original fractions square to 0. Always combine reciprocal sums properly to find the pairwise product sum. ### Final Answer Therefore, the correct answer is **1**.