A weird system of equations.

From the first equation we get that $x, y, z <\frac{1}{2}$ .

$x^{2}+y^{2}+z^{2}=\frac{3}{2} \times \frac{1}{6} = \frac{3}{2} (x^{3}+y^{3}+z^{3})$

Therefore $x^{2}(1-\frac{3}{2}x)+y^{2}(1-\frac{3}{2}y)+z^{2}(1-\frac{3}{2}z)=0$

But $1-\frac{3}{2}x> 1-\frac{3}{4} = \frac{1}{4}$

Therefore $x^{2}(1-\frac{3}{2}x)$ is greater than $0$ or equal to $0$ but $x^{2}(1-\frac{3}{2}x)+y^{2}(1-\frac{3}{2}y)+z^{2}(1-\frac{3}{2}z)=0$ so $x^{2}(1-\frac{3}{2}x)=0$ and this means $x=0, y=0, z=0$ which is not a solution.

Taking the double of each unknown shows that the sum of their sqares is 1, while the sum of the cubes is 8/6, i. e. greater than 1. This is absurd, since a higher power of a number below unity is smaller.

To build on Paul Gartner's solution:

Consider the transformation $(t,u,v) = 2*(x,y,z)$ . Then $t^2 + u^2 + v^2 = 1$ and $t^3 + u^3 + v^3 = \frac{4}{3}.$ The first equation here implies that all the variables are of magnitude at most 1, and thus $|t|^3 \leq |t|^2$ , and $|u|^3 \leq |u|^2$ and $|v|^3 \leq|v|^2$ .

This gives us the chain of inequalities:

$|t^3 + u^3 + v^3| \leq |t^3| + |u^3| + |v^3| \leq |t|^2 + |u|^2 + |v|^2 = t^2 + u^2 + v^2 = 1$

The first inequality is because $|a + b| \leq |a| + |b|$ for all real numbers $a,b$ . The second inequality is because the magnitudes of all the variables is bounded by 1. The absolute value signs may be removed because squaring the numbers makes them all non-negative.

In conclusion $|t^3 + u^3 +v^3| \leq 1 < 4/3$ , meaning no solution is possible.

Suppose, for contradiction, that $v=(x,y,z) \in R^{3}$ such that the system is satisfied by x,y,z. Then, $||v||_{2} = 4^{-\frac{1}{2}} = \frac{1}{2}, ||v||_{3} = 6^{-\frac{1}{3}}$ by the definition of the $L^{p}$ norms. Hence, $||v||_{3}\leq ||v||_{2}$ by the basic properties of $L^{p}$ norms. On the other hand, $6^{-\frac{1}{3}} > \frac{1}{2}$ . This is a contradiction. Hence, no such vector solves the system; the system has no solutions.

$x^{3}+y^{3}+z^{3}=\frac{1}{6}$

$x^{2}+y^{2}+z^{2}=\frac{1}{4}$

Multiply the first equation by $\frac{3}{2}$ to get $\frac{3}{2}(x^{3}+y^{3}+z^{3})=\frac{1}{4}$ .

So $x^{2}+y^{2}+z^{2}=\frac{3}{2}(x^{3}+y^{3}+z^{3})=\frac{1}{4}$ .

Now rewrite like $\frac{3}{2}x\times x^{2}+\frac{3}{2}y\times y^{2}+\frac{3}{2}z\times z^{2}=x^{2}+y^{2}+z^{2}$ .

We see that $x=y=z=\frac{2}{3}$ which does not satisfy the system.

Wolfram Alpha

Shows no real roots

Let {x = a, y = b, z = c} by a solution. Then { x = a, y = c, z = b} would also be a solution. This implies the number of triplets be a multiple of 3!. The only possible answer given the choices is 0.

They are a pair of concentric spheres with different radios, so, there is not interception.

My fault,I read the coefficients in a wrong way. Downvote to myself

We can solve it by it's geometric meaning. So, the answer is just the number of these 2 graphs' intersections.

The first equation stands for a spherical surface, whose radius equals $\frac{1}{2}$ , it's highly symmetrical. :)

The second equation is also symmetrical, which means, if you found a solution/point $(x_{0}, y_{0}, z_{0})$ in octant I (where $x>0, y>0 and z>0$ ) , just rotate it 2 times through the line $x=y=z$ , and you will get another 2 solutions, they are $(y_{0}, z_{0}, x_{0})$ and $(z_{0}, x_{0}, y_{0})$ . (only when that point does not belong to the line).

So the answer can be divided by 3! without remainder.

Assume that $x = y = z = \frac{\sqrt{3}}{6}$ , this does not satisfy the second equation.

The only choice is zero. XD

In a 3D graphing program, solid modeller or other such, create the 0.5 radius sphere. This is the set of points which solver the first equation. Then model the surface over the range |x|<0.5 and |y|<0.5 of z = (1/6 - x^3 -y^3)^(1/3). Then explore in 3D. You can see the surface never gets to touch or intersect the sphere.

By graphing both equations, you can see that there is no intersection. Check it out!

First equation is of eqa of sphere of radius 1/2 . To have a solution of given system of equations, we need second curve should cut the sphere at some place. From second equation, highest value of x(y=z=0) = cube root of 1/6 << 1/2 radius of sphere. So, there is no real solution for this system of equations

This solution it's acutally simple, i just didn't skiped any step.

$x^{2} + y^{2} + z^{3} = \frac{1}{4}$

$x^{3} + y^{3} + z^{3} = \frac{1}{6}$

Sum up two equations and we end with $RHS = 5/12$

Then, i put in evidence the common factors $x^2 , y^2, z^2$ (on $LHS)$

$x^2(1+x) + y^2(1+y) + z^2(1+z) = \frac{5}{12} = \frac{1}{4} \times \frac{5}{3}= (x^2 + y^2 + z^2)(\frac{5}{3}) = x^2\frac{5}{3} + y^2\frac{5}{3} + z^2\frac{5}{3}$

now looking by the simmetry of the equation we do have that

$(1 +x)=(1+y)=(1+z)= \frac{5}{3}$

So

$x=y=z$

But there's no solution for the system :

$3x^2 = \frac{1}{4}$

$3x^3 = \frac{1}{6}$

So the actual answer is $0$ ... not $1$ as we first get by looking at $x=y=z=\frac{2}{3}$

Probably not recommended, but since first equation is sum of squares, each positive solution can be replaced with negative. This implies an even number of solution.

Here's my approach: Every system of equations with n variables needs n variables to be solved. If this is not the case, the system has either infinite solutions or no solutions.