Is the system uniquely solvable?

Question

Is the system $$x^3+y^3+z^3=1 \\ x\cdot y\cdot z=-1$$ in a neighbourhood of the point $(1; -1; 1)$ uniquely solvable for $y = y (x) $ and $z = z (x)$ ?

$$$$

According to that theorem we have the following:

We have the continuous differentiable function $F(x,y,z)=x^3+y^3+z^3-1$and the point $(1,-1,1)$ with $F(1,-1,1)=0$. We have that $\frac{\partial{F}}{\partial{z}}=3z^2\Rightarrow \frac{\partial{F}}{\partial{z}}(1,-1,1)=3\neq 0$. Then there is a neighborhood $(1,-1,1)$ so that whenever $(x, y)$ is sufficiently close to $(x_0, y_0)$ there is a unique $z$ so that $F(x, y, z) = 0$. Moreover, this assignment is makes $z$ a continuous function of $x$ and $y$.

Do we get from that the desired result?

Answer 1

Your question seems to ask if you can write $y$ and $z$ depending on $x$ and not $z$ depending on $x$ and $y$. Moreover, you just considered the first equation but drops the second.

You should consider $$ F(x,y,z)=\begin{pmatrix}x^3+y^3+z^3-1\\xyz-1\end{pmatrix}. $$ Check $F(1,-1,1)$ and $\partial_{y,z}F(1,-1,1)$. Then you can apply the implicite function theorem to get open neighbourhoods $U$ of $1$ and $V$ of $(-1,1)$ and a differentiable bijection $$ g:U\to V $$ such that $$ F(x,g_1(x),g_2(x))=0~\forall x\in U, $$ where $g_i(x)$ is the $i$.th component of $g(x)$. You could say $y(x)=g_1(x)$ and $z(x)=g_2(x)$. The function $g$ as it component functions is as smooth as $F$ is.

Answer 2

Eliminating $y$, we get $x^3 z^6 + (x^6-x^3) z^3 - 1 = 0$, which is a quadratic in $z^3$ with discriminant $x^{12} - 2 x^9 + x^6 + 4 x^3$. This is strictly positive for $x$ in a neighbourhood of $1$, so there are two distinct real roots in that neighbourhood (one of which will be near $z=1$ and the other near $z=-1$).

Answer 3

Let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative, and $xyz=w^3$.

Hence, $x^3+y^3+z^3=1$ gives $$27u^3-27uv^2+3w^3=1$$ or $$v^2=u^2-\frac{4}{27u}.$$ In another hand, $$(x-y)^2(x-z)^2(y-z)^2\geq0$$ gives $$3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6\geq0$$ or $$3u^2\left(u^2-\frac{4}{27u}\right)^2-4\left(u^2-\frac{4}{27u}\right)^3+4u^3-6u\left(u^2-\frac{4}{27u}\right)-1\geq0$$ or $$\frac{(3u-1)(9u^2+3u+1)(729u^6+837u^3+256)}{u^3}\leq0$$ or $$0<u\leq\frac{1}{3},$$ which gives infinitely many solutions.