Determine all triples of real numbers $(a,b,c)$ such that \begin{eqnarray*} xyz &=& 8 \\ x^2y + y^2z + z^2x &=& 73 \\ x(y-z)^2 + y(z-x)^2 + z(x-y)^2 &=& 98 . \end{eqnarray*}
Problem
Source: Iberoamerican 2005
Tags: quadratics, algebra unsolved, algebra
manuel
29.09.2005 01:18
Expanding the third equation we get $xy^2-2xy+z^2x+yz^2-2xyz+yx^2+zx^2-2xy+y^2z=98$ Now by substituing the first two equations we get: $xy^2+yz^2+zx^2=73$ and by the second equation we obtain that $xy^2+yz^2+zx^2=x^2y+y^2z+z^2x$. So since this eq. is symmetric and we can't have $x>y>z$, two of them must be equal, say $x=y$. So we have from the first eq. $x^2z=8$. Plugging this and $x=y$ in the second eq. we have that $x^3+64/(x^3)-65=0$, So put $x^3=a$, and we get $a^2-65a+64=0$. Solving this quadratic we get that $a=64$ or $a=1$, therefore $x=4$ or $x=1$. So the solutions are $(1,1,8)$ and $(1/2,4,4)$, and of course all the permutations of these.
Pascual2005
02.10.2005 22:51
$(x-y)(y-z)(z-x)=xy^2+yz^2+zx^2-x^2y-y^2z-z^2x$ so $x=y$ or $y=z$ or $z=x$
RMACR7LP
02.01.2018 21:36
Expanding the third equation we get
\[xy^2+yz^2+zx^2-6xyz+xz^2+yx^2+zy^2=98\]substituting the first two equations we achieve $xy^2+yz^2+zx^2=xz^2+yx^2+zy^2 \rightarrow xy(y-x)+yz(z-y)+xz(x-z)=0$ Substituting the first equation again;
\[\frac{8(y-x)}{z}+\frac{8(z-y)}{x}=xz(z-x)\](remark: this equation is well defined since $z\neq 0$ and $x\neq0$.)
Multiplying the equation by $xz$ and rearranging the LHS;
\[8y(x-z)+8(z-x)(z+x)=x^2z^2(z-x)\]Now we put all the terms to one side of the equation and factor $z-x$ resulting in $z=x$ or $x^2z^2+8y-8(z+x)=0$.
If $x^2z^2+8y-8(z+x)=0$ then $x^2y^2=8(x+z-y)$ and by the first equation, $xz=y(x+z-y)$ finally put all in one side and factor ; $(x-y)(z-y)=0$. We then know that at least two variables are the same.
WLOG $z=x$
The first equation becomes $x^2y=8$, the second $x^2y+xy^2+x^3=73$, and third equation $x(x-y)^2=49$. Substituting the first in the second and then plugging that into the third equation results in
\[8+\frac{64}{x^3}+x^3=73\]which is a quadratic equation. Solving we see that the solutions are $(4, 1/2,4), (1,8,1)$ and their permutations.