STEP III 2009 5

5

\begin{align} x+y+z=1\implies(x+y+z)^2 &=((x+y)+z)^2 \\&= (x+y)^2+2(x+y)z+z^2 \\&= x^2+y^2+z^2+2(xy+yz+zx)=1^2=1 \\ \iff xy+yz+zx&=\frac{1}{2}(1-(x^2+y^2+z^2)) \\&= \frac 1 2 (1-2)=-\frac 1 2 \end{align}
\begin{align} x^2y+x^2z+y^2z+y^2x+z^2x+z^2y&=x^2(y+z)+y^2(x+z)+z^2(x+y) \\&=x^2(1-x)+y^2(1-y)+z^2(1-z) \\&=(x^2+y^2+z^2)-(x^3+y^3+z^3) \\&=2-3=-1 \end{align}
\begin{align} (x+y+z)^3 &= (x+y+z)((x+y)+z)^2 \\&= (x+y+z) (x^2+y^2+z^2+2(xy+yz+zx)) \\&= x^3+yx^2+zx^2+xy^2+y^3+zy^2+xz^2+yz^2+z^3+2(xy+yz+xz)(x+y+z) \\&= x^3+yx^2+zx^2+xy^2+y^3+zy^2+xz^2+yz^2+z^3+2(x^2y+xy^2+xyz+xyz+y^2z+yz^2+x^2z+xyz+xz^2) \\&= (x^3+y^3+z^3)+3(x^2y+x^2z+y^2z+y^2x+xz^2+yz^2)+6xyz \\&= 3+3(-1)+6xyz \\\implies (1)^3 &= 6xyz \\\iff xyz &= \dfrac 1 6 \end{align}
\begin{align} &S_{n+1}=aS_n+bS_{n-1}+cS_{n-2} \\ &\iff x^{n+1}+y^{n+1}+z^{n+1}=a(x^n+y^n+z^n)+b(x^{n-1}+y^{n-1}+z^{n-1})+c(x^{n-2}+y^{n-2}+z^{n-2}) \\ &\iff (x^{n+1}-ax^n-bx^{n-1}-cx^{n-2})+(y^{n+1}-ay^n-by^{n-1}-cy^{n-2})+(z^{n+1}-az^n-bz^{n-1}-cz^{n-2})=0 \\ &\iff x^{n-2}(x^3-ax^2-bx-c)+y^{n-2}(y^3-ay^2-by-c)+z^{n-2}(z^3-az^2-bz-c)=0 \end{align}

define

f(n)=n^3-an^2-bn-c

so that x,y,z are roots of f.
but if they are the roots of f then by vieta’s formula, the results above give the coefficients of f, namely a,b,c.
we know

\begin{align} x+y+z&=1 \\ xy+yz+zx&=-\dfrac{1}{2} \\ xyz &=\dfrac{1}{6} \\ \implies a &= 1 \\ b&= -\dfrac{1}{2} \\ c&= -\dfrac{1}{6} \end{align}
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