There is a problem:

Is there any triple of rational numbers $(a,b,c)$ such that $a+b+c=0$ and $abc=1$?

My answer is $\boxed{No}$. Here is my proof:

Suppose such triple exists. Then they must be the $3$ solutions (counting multiplicities) of the cubic equation: $$X^3+kX-1=0,$$ where $k$ is a rational number and $k\ne 0$ (because $k=0$ makes the equation get non-rational solutions). This equation has two roots: $$X_{1,2}=\frac{1}{2}\pm \sqrt{\frac{1}{4}+\frac{k^3}{27}}.$$ Thus $\frac{1}{4}+\frac{k^3}{27}=m^2$ with some rational number $m$. By a linear change of variables, the result is equivalent to the Elliptic curve $$(E): y^2=x^3+\frac{1}{4}$$ having a non-trivial rational point. But $(E)$ has rank $0$ and torsion group $\mathbb{Z}/3\mathbb{Z}$ (checked by CoCalc) so $E(\mathbb{Q})=\{O, (0,\frac{1}{2}),(0,-\frac{1}{2})\}$, a contradiction. This proved the claim.

Did I make a mistake? I am still looking for an elementary solution.

There is a problem:

Is there any triple of rational numbers $(a,b,c)$ such that $a+b+c=0$ and $abc=1$?

My answer is $\boxed{No}$. Here is my proof:

Suppose such a triple exists. Then it must consist of the $3$ solutions (counting multiplicities) of the cubic equation: $$X^3+kX-1=0,$$ where $k$ is a rational number and $k\ne 0$ (because $k=0$ makes the equation get non-rational solutions). This equation has two roots: $$X_{1,2}=\frac{1}{2}\pm \sqrt{\frac{1}{4}+\frac{k^3}{27}}.$$ Thus $\frac{1}{4}+\frac{k^3}{27}=m^2$ with some rational number $m$. By a linear change of variables, the result is equivalent to the Elliptic curve $$(E): y^2=x^3+\frac{1}{4}$$ having a non-trivial rational point. But $(E)$ has rank $0$ and torsion group $\mathbb{Z}/3\mathbb{Z}$ (checked by CoCalc) so $E(\mathbb{Q})=\{O, (0,\frac{1}{2}),(0,-\frac{1}{2})\}$, a contradiction. This proved the claim.

Did I make a mistake? I am still looking for an elementary solution.