Newton says Polynomial

As you know identity
$r^3+s^3+t^3-3rst=(r+s+t)(r^2+s^2+t^2-rs-st-tr)$
→Now, transposing $(-3rst)$ to RHS .
$r^3+s^3+t^3=(r+s+t)(r^2+s^2+t^2-rs-st-tr)+3rst$ ....... $(★)$
Now applying Vieta's formula in cubic polynomial $x^3+3x^2+4x-8$ .
Taking Roots as $r,s,t$ .
● $r+s+t=\frac{-b}{a}=-3$ ..... $(1)$
● $rs+st+tr=\frac{c}{a}=4$ ..... $(2)$
● $r×s×t=\frac{-d}{a}=8$ ...... $(3)$
Now, Simplying $(★)$
$r^3+s^3+t^3=(r+s+t)(r^2+s^2+t^2-rs-st-tr)+3rst=(r+s+t)[(r+s+t)^2-3(rs+st+tr)]+3rst$
Now, putting values from $(1),(2),(3)$
$r^3+s^3+t^3=(-3)[(-3)^2-3(4)]+3×8$
$r^3+s^3+t^3=(-3)×(-3)+24$
$r^3+s^3+t^3=\boxed{33}$

Let $P(x)=x^3+3x^2+4x-8$ and $P_{n}=r^{n}+s^{n}+t^{n}$ .

By using $\text{Newton's Sums}$ ,

$(1)P_{1}+3=0\Rightarrow P_{1}=-3 \\ (1)P_{2}+(3)P_{1}+8=0\Rightarrow P_{2}=1 \\ (1)P_{3}+(3)P_{2}+(4)P_{1}-24=0\Rightarrow P_{3}=\boxed{33}$

Using the identity:

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

Putting the values from the $P(x)$

$\Rightarrow$ $r^{3}+s^{3}+t^{3}=(r+s+t)(r^{2}+s^{2}+t^{2}-rs-st-tr)+3rst$

$\Rightarrow$ $r^{3}+s^{3}+t^{3}=(-3)(-3)+24=33$

We have,
x³+3x²+4x-8=0.
x³-x²+4x²-4x+8x-8=0.
x²(x-1)+4x(x-1)+8(x-1)=0.
(x-1)(x²+4x+1)=0.
Therefore, r=1.
Now, for x²+4x+1=0.
s+t=-4 and st=8.
Using these two equations,
s³+t³=32.
Therefore, r³+s³+t³=33.