Mathematics — CBSE Class 10 board question

For $px^2 + qx + r$, using the relations between zeroes and coefficients:

$$\alpha + \beta = \frac{-q}{p}, \qquad \alpha\beta = \frac{r}{p}$$

Now, $\alpha^3\beta + \beta^3\alpha = \alpha\beta(\alpha^2 + \beta^2) = \alpha\beta\,[(\alpha+\beta)^2 - 2\alpha\beta]$

$$= \frac{r}{p}\left[\frac{q^2}{p^2} - \frac{2r}{p}\right] = \frac{r}{p} \cdot \frac{q^2 - 2rp}{p^2} = \frac{r(q^2 - 2pr)}{p^3}$$

Source: Chapter 2, Section 2.3 — Relationship between Zeroes and Coefficients of a Polynomial

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• First, write the standard sum and product of zeroes for $px^2+qx+r$.
• The key algebraic step is factoring: $\alpha^3\beta + \alpha\beta^3 = \alpha\beta(\alpha^2+\beta^2)$, then use $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$ to express everything in terms of $p, q, r$.
• Examiners award marks for the factoring step and the correct substitution — show both clearly.