Question

If \( \alpha, \beta, y \) are the roots of \( x^{3}-2 x^{2}+3 x-4=0 \) then \( \alpha^{2} \beta^{2}+\beta^{2} \gamma^{2}+\gamma^{2} \alpha^{2} \) is

Answer 1

\(\alpha , \beta \) and \(\gamma\) are roots of \(x^3 - 2x^2 + 3x - 4 = 0\) 

then sum of roots = \(\alpha + \beta + \gamma = \frac{-b}a = \frac{-(-2)}1 = 2\)

\(\alpha \beta + \beta \gamma+\alpha\gamma = \frac ca = \frac 31=3\)

\(\alpha \beta \gamma = \frac{-d}a = \frac{-(-4)}1 = 4\)

Now, \((\alpha \beta + \beta \gamma + \alpha\gamma)^2 = \alpha ^2\beta^2 + \beta^2\gamma ^2 + \alpha^2\gamma^2 + 2\alpha\beta^2\gamma+ 2\alpha^2\beta\gamma + 2\alpha\beta\gamma^2\)

⇒ \(\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 + 2\alpha\beta \gamma(\alpha + \beta) = (\alpha \beta + \beta \gamma+ \alpha\gamma)\)

⇒ \(\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2 \alpha^2 + 2 \times 4\times 2 = 3^2\)

⇒ \(\alpha^2 \beta^2+ \beta^2\gamma^2 + \gamma^2 \alpha^2 = 9 - 16 \)

⇒ \(\alpha^2 \beta^2+ \beta^2\gamma^2 + \gamma^2 \alpha^2 = -7\)