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For this quadratic equation,
$\text { Let } \alpha, \alpha^{2} \text { be the roots of } 3 x^{2}+p x+3=0$
$\text { Now, } \quad S=\alpha+\alpha^{2}=-p / 3 \text {, }$
$P=\alpha^{3}=1$
$\Rightarrow \quad \alpha=1, \omega, \omega^{2}$
$\text { Now, } \quad \alpha+\alpha^{2}=-p / 3$
$\Rightarrow \omega+\omega^{2}=-p / 3$
$\Rightarrow -1=-p / 3$
$\Rightarrow p=3$
Hence p = 3.
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