$p\left(x\right)=\left(3x^3–10x^2–27x+10\right)$
p(x)=\left(3 x^{3}-10 x^{2}-27 x+10\right)
$p\left(5\right)=\left(3\times 5^3–10\times 5^2–27\times 5+10\right)=(375–250–135+10)=0$
p(5)=\left(3 \times 5^{3}-10 \times 5^{2}-27 \times 5+10\right)=(375-250-135+10)=0
$p(–2)=\left[3\times \left(–2^3\right)–10\times \left(–2^2\right)–27\times (–2)+10\right]=(–24–40+54+10)=0$
p(-2)=\left[3 \times\left(-2^{3}\right)-10 \times\left(-2^{2}\right)-27 \times(-2)+10\right]=(-24-40+54+10)=0
$\mathrm{p}\left(\frac{1}{3}\right)=\left\{3\times {\left(\frac{1}{3}\right)}^3–10\times {\left(\frac{1}{3}\right)}^2–27\times \frac{1}{3}+10\right\}=\left(3\times \frac{1}{27}–10\times \frac{1}{9}–9+10\right)$
\mathrm{p}\left(\frac{1}{3}\right)=\left\{3 \times\left(\frac{1}{3}\right)^{3}-10 \times\left(\frac{1}{3}\right)^{2}-27 \times \frac{1}{3}+10\right\}=\left(3 \times \frac{1}{27}-10 \times \frac{1}{9}-9+10\right)
$=\left(\frac{1}{9}–\frac{10}{9}+1\right)=\left(\frac{1–10–9}{9}\right)=\left(\frac{0}{9}\right)=0$
=\left(\frac{1}{9}-\frac{10}{9}+1\right)=\left(\frac{1-10-9}{9}\right)=\left(\frac{0}{9}\right)=0
$\therefore 5,–2\text{ and }\frac{1}{3}\text{ are the zeroes of }p\left(x\right)\text{. }$
\therefore 5,-2 \text { and } \frac{1}{3} \text { are the zeroes of } p(x) \text {. }

Let $\alpha=5$, $\beta=-2$ and $\gamma=\frac{1}{3}$. Then we have:

$(\alpha+\beta+\gamma)=\left(5-2+\frac{1}{3}\right)=\frac{10}{3}=\frac{-\left(\text { coef ficient of } x^{2}\right)}{\left(\text { coefficient of } x^{3}\right)}$

$(\alpha \beta+\beta \gamma+\gamma \alpha)=\left(-10-\frac{2}{3}+\frac{5}{3}\right)=\frac{-27}{3}=\frac{\text { coefficient of } x}{\text { coefficient of } x^{3}}$

$\alpha \beta \gamma=\left\{5 \times(-2) \times \frac{1}{3}\right\}=\frac{-10}{3}=\frac{-(\text { constant term })}{\left(\text { coef ficient of } x^{3}\right)}$