Question

If $\large\rm { \alpha , \beta , \gamma}$ are roots of $\large\rm { x^{3} + ax^{2} + b =0}$ then the value of determinant $\left[\ \textless \ br /\ \textgreater \ \begin{array}{c c c}\ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ \alpha & \beta & \gamma \\\ \textless \ br /\ \textgreater \ \beta & \gamma & \alpha \\\ \textless \ br /\ \textgreater \ \gamma & \alpha & \beta \ \textless \ br /\ \textgreater \ \end{array}\ \textless \ br /\ \textgreater \ \right]$​

Answer 1

♣ Qᴜᴇꜱᴛɪᴏɴ :

If α, β, γ are the roots of x³ + ax² + b = 0, then the value of determinant

$\left|\begin{array}{lll}\alpha & \beta & \gamma \\\beta & \gamma & \alpha \\\gamma & \alpha & \beta\end{array}\right|$

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♣ ᴀɴꜱᴡᴇʀ :

$\huge\boxed{\sf{\mathrm{a}^{3}}}$

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♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

Because α, β, γ are the roots of x³ + ax² + b = 0, therefore,

$\sf{\alpha+\beta+\gamma=-a, \alpha \beta+\beta \gamma+\gamma \alpha=0, \alpha \beta \gamma=-\beta}$

$\left|\begin{array}{ccc}\alpha & \beta & \gamma \\\beta & \gamma & \alpha \\\gamma & \alpha & \beta\end{array}\right|=-(\alpha+\beta+\gamma)\left[\alpha^{2}+\beta^{2}\right.\left.+\gamma^{2}-\alpha \beta-\beta \gamma-\gamma \alpha\right]$

= -(-a)[(-a)² - 3 × 0]

= a × a²

= a³

Alternative method :

$\sf{\text { Value of determinant }=\Delta=\alpha\left(\beta \cdot \gamma-\alpha^{2}\right)-\beta\left(\alpha \cdot \gamma-\beta^{2}\right)+\gamma\left(\alpha \cdot \beta-\gamma^{2}\right)}$

$\begin{array}{l}\rightarrow \Delta=3 . \alpha . \beta \cdot \gamma-\left(\alpha^{3}+\beta^{3}+\gamma^{3}\right) \\\\\rightarrow \alpha . \beta \cdot \gamma=-\mathrm{b} \\\\\rightarrow \alpha+\beta+\gamma=-\mathrm{a} \\\\\rightarrow \alpha \beta+\beta \gamma+\gamma \alpha=0\end{array}$

$\begin{array}{l}\text { And } \alpha^{3}+\beta^{3}+\gamma^{3}-3 \alpha \beta \gamma=(\alpha+\beta+\gamma)\left(\alpha^{2}+\beta^{2}+\gamma^{2}-\alpha \beta-\beta \gamma-\gamma \alpha\right) \\\\\text { Also } \alpha^{2}+\beta^{2}+\gamma^{2}=(\alpha+\beta+\gamma)^{2}-2(\alpha \beta+\beta \gamma+\gamma \alpha)\end{array}$

Substituting we get ;

$\begin{array}{l}\alpha^{3}+\beta^{3}+\gamma^{3}=3(-\mathrm{b})+(-\mathrm{a})\left(\mathrm{a}^{2}\right) \\\\\therefore \Delta=\mathrm{a}^{3}+3 \mathrm{b}-3 \mathrm{b}\end{array}$

$\large\boxed{\sf{\Rightarrow \Delta=\mathrm{a}^{3}}}$

Answer 2

Step-by-step explanation:

Hope it helps u........