Problem source

\begin{questionparts}
	 \item If $x+y+z=\alpha,$ $xy+yz+zx=\beta$ and $xyz=\gamma,$ find numbers $A,B$ and $C$ such that 
\[
x^{3}+y^{3}+z^{3}=A\alpha^{3}+B\alpha\beta+C\gamma.
\]
Solve the equations 
\begin{alignat*}{1}
x+y+z & =1\\
x^{2}+y^{2}+z^{2} & =3\\
x^{3}+y^{3}+z^{3} & =4.
\end{alignat*}
\item The area of a triangle whose sides are $a,b$ and $c$ is given by the formula 
\[
\mathrm{area}=\sqrt{s(s-a)(s-b)(s-c)}
\]
where $s$ is the semi-perimeter $\frac{1}{2}(a+b+c).$ If $a,b$ and $c$ are the roots of the equation 
\[
x^{3}-16x^{2}+81x-128=0,
\]
find the area of the triangle. 
\end{questionparts}

Solution source

\begin{questionparts}
\item \begin{align*}
(x+y+z)^3 &= x^3+y^3+z^3+ \\
&\quad 3xy^2 + 3xz^2 + 3yx^2 + \cdots + 3zy^2 \\
&\quad\quad + 6xyz \\
(x+y+z)(xy+yz+zx) &= x^2y+x^2z + \cdots + z^2 x + 3xyz \\
x^3+y^3+z^3 &= (x+y+z)^3 - 3(xy^2 + \cdots + zy^2) - 6xyz \\
&= \alpha^3 - 3(\alpha \beta - 3\gamma)-6\gamma \\
&= \alpha^3-3\alpha \beta+3\gamma
\end{align*}

Since $4 = 1^3-3\cdot1\cdot(-1) + 3 \gamma \Rightarrow \gamma = 0$, therefore one of $x,y,z = 0$. WLOG $x = 0$, so 

$y+z = 1, y^2 + z^2 = 3 \Rightarrow y^2 + (1-y)^2 = 3 \Rightarrow y^2 -y -1 = 0 \Rightarrow y = \frac{1 \pm \sqrt{5}}{2}$, so we have

$(x,y,z) = (0, \frac{1 +\sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2})$ and permutations.

\item \begin{align*}
A^2 &= s(s-a)(s-b)(s-c) \\
\end{align*}

Notice the second part is the same as plugging $s= 16/2 = 8$ into our polynomial

Therefore 
\begin{align*}
A^2 &= 8 \cdot (8^3 - 16 \cdot 8^2 + 81 \cdot 8 - 128) \\
&= 8 \cdot 8 (8^2 - 16 \cdot 8 + 81- 16)  \\
&= 64 (-64+81-16) \\
&= 64
\end{align*}

Therefore $A = 8$
\end{questionparts}