Does $x^3+y^3=\frac{1}{\sqrt2}$ have points inside $(x,y): x^2+y^2 \leq 1$ How can I tell that the curve $x^3+y^3=\frac{1}{\sqrt2}$ have points on the compact set $(x,y): x^2+y^2 \leq 1$ ? 
Can I tell somehow right away, instead of solving the system of equations?
 A: You can tell right away by guessing. Simply setting one of the variables to $0$ reveals that $(2^{-1/6},0)$ is on the cubic curve and in the disc. Setting $x=y$ also reveals that $(1/\sqrt2,1/\sqrt2)$ is on the cubic and on the disc boundary.
A: It should be clear that the point $\left(\frac1{\sqrt2},\frac1{\sqrt2}\right)$ belongs to both sets.
A: One of the points on $x^3+y^3=\frac{1}{\sqrt2}$ is $\left(\frac{1}{\sqrt[6]2},0\right)$
Since $0 \lt \frac{1}{\sqrt[6]2} \lt 1$, you can say this point satisfies $x^2+y^2 \leq 1$
A: Part of the curve  $$ x^3+y^3= \frac {1}{\sqrt 2}$$ goes true  the disk $x^2+y^2=1$ and the intersection is uncountable.
The intersection passes through the point $(0, 2^{-1/6})$ which is inside the disk.   
A: Given a curve of the type $$x^3+y^3=a,$$ with $a$ real, we may write this as $$y={(a-x^3)}^{1/3},$$ which defines $y$ as an everywhere continuous function of $x.$ Furthermore, as $x\to\pm\infty,$ we have that $y\to\mp\infty.$ Thus, the curve divides the plane into two regions. Furthermore, we have $y= a^{1/3}$ when $x=0.$ Thus, we only need to have $$-1\le a^{1/3}\le 1$$ in order to have some point of the cubic within the closed unit disk.
If $a=\frac{1}{\sqrt 2},$ then $|a^{1/3}|$ is indeed less than unity since $2^{1/6}>1,$ which is true since $2^{1/2}>1.$ By the continuity of the involved curves it follows that there are infinitely many points of the cubic not without the disk.
