$ \frac{\alpha^2}{-4k-15}=\frac{\alpha}{-5+8}=\frac{1}{-6-k}$
$ \frac{\alpha^2}{-4k-15}=\frac{\alpha}{3}=\frac{1}{-6-k}$
$ \alpha^2=\frac{4k+15}{k+6},\alpha=\frac{-3}{k+6}$
$ \Rightarrow \alpha^2=\alpha^2$
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$\alpha$ be the common root
$ \Rightarrow a \alpha^2+b \alpha+c=0 \longrightarrow (1) $
$ \Rightarrow b \alpha^2+c \alpha+a=0 \longrightarrow(2)$
$ \frac{\alpha^2}{a b-c^2}=\frac{\alpha}{c b-a^2}=\frac{1}{a c-b^2} $
$\alpha^2=\frac{a b-c^2}{a c-b^2} $
$\alpha=\frac{c b-a^2}{a c-b^2} $
$ \Rightarrow \alpha^2=(\alpha)^2$
$ \Rightarrow \frac{ab-c^2}{ac-b^2}=(\frac{cb-a^2}{ac-b^2})^2$
$ \Rightarrow (ab-c^2)(ac-b^2)=(cb-a^2)^2$
$ \Rightarrow a^2bc-ab^3-ac^3+b^2c^2 = b^2c^2+a^4-2a^2bc$
$\Rightarrow 3 a^2 b c=a^4+a b^3+a c^3$
$ \Rightarrow a^3+b^3+c^3=3abc$
$ a_0x^3+a_1x^2+a_2^x+a_3=0$
$ a_0 \neq0 \ \ ,a_0,a_1,a_2,a_3 \epsilon R$
$\rArr$ Relation between roots of cubic equation and its coefficient.
$ \Rightarrow a_0x^3+a_1x^2+a_2x+a_3=0 $
$ \alpha,\beta,\gamma \to roots $
$ \Rightarrow a_0x^3+a_1x^2+a_2x+a_3=a_0(x-\alpha)(x-\beta)(x-\gamma)$
$ \Rightarrow a_0x^3+a_1x^2+a_2x+a_3=a_0[(x^2-(\alpha+\beta)x+\alpha\beta] (x-\gamma)$
$ \Rightarrow a_0x^3+a_1x^2+a_2x+a_3=a_0[(x^3-\gamma x^2-(\alpha+\beta)x^2+(\alpha+\beta).\gamma x+\alpha\beta x-\alpha\beta\gamma]$
$=a_0[(x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\gamma\alpha)x-\alpha\beta\gamma]$
$ \Rightarrow a_0x^3+a_1x^2+a_2x+a_3=a_0x^3-a_0(\alpha+\beta+\gamma)x^2+a_0(\alpha\beta+\beta\gamma+\gamma\alpha)x-a_0 \alpha\beta\gamma$
$ \Rightarrow x^2 \to $
$ a_1=-a_0(\alpha+\beta+\gamma)$
$ S_1=\alpha+\beta+\gamma=-\frac{a_1}{a_0}$
$x \to a_2=a_0(\alpha\beta+\beta\gamma+\gamma\alpha)$
$ S_2=\alpha\beta+\beta\gamma+\gamma\alpha= \frac{a_2}{a_2}$
$ conc.\ S_3=\alpha\beta\gamma=-\frac{a_3}{a_0}$
Find the value of $\alpha+\beta+\gamma$, $\alpha\beta+\beta\gamma+\gamma\alpha$ and $\alpha\beta\gamma$ if $\alpha,\beta,\gamma$ are the roots of cubic equation $x^3+6x^2+5x-12=0$
$ sol: a_0x^3+a_1x^2+a_2x+a_3=0$
$ a_0=1, a_1=6,a_2=5,a_3=-12$
$ \alpha+\beta+\gamma=-\frac{a_1}{a_0}$
$ =-\frac{6}{1} $
$ =-6 $
$ \alpha\beta+\beta\gamma+\gamma\alpha=-\frac{a_2}{a_0}$
$ =\frac{5}{1} $
$ =5 $
$ \alpha\beta+\beta\gamma+\gamma\alpha=-\frac{a_3}{a_0}$
$ =+\frac{12}{1} $
$ =12 $