UPSC NDA I 2022 Mathematics Solved Paper

Concept:For a quadratic equation $x^2 + px + q = 0$ with roots $\alpha$ and $\beta$,the sum of roots is $\alpha + \beta = -p$, and the product is $\alpha \beta = q$.Similarly, for $x^2 + mx + n = 0$ with roots $\alpha^3$ and $\beta^3$,sum $\alpha^3 + \beta^3 = -m$ and product $\alpha^3 \beta^3 = n$.We use the identity $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$.Explanation:From the first equation: $\alpha + \beta = -p$ and $\alpha\beta = q$.Compute $\alpha^3 + \beta^3$ using the identity:$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$.Substitute the values: $\alpha^3 + \beta^3 = (-p)^3 - 3(q)(-p) = -p^3 + 3pq$.For the second equation, the sum of roots is $\alpha^3 + \beta^3 = -m$.Thus $-m = -p^3 + 3pq$, so $m = p^3 - 3pq$.The product of roots is $\alpha^3\beta^3 = (\alpha\beta)^3 = q^3 = n$.Hence $n = q^3$.Therefore $m + n = (p^3 - 3pq) + q^3 = p^3 + q^3 - 3pq$.Answer:$m + n = p^3 + q^3 - 3pq$ (Option D).