I continue beating the dead horse that is this problem by exhibiting a solution in terms of trigonometric functions.
Given that ##1=a^{2}+b^{2}+c^{2}##, it's natural to interpret ##\left(a,b,c\right)## as the components of a unit-vector in 3D and parametrize those three components in terms of two spherical angles ##\theta,\phi\,##:$$a=\sin\theta\cos\phi,\;b=\sin\theta\sin\phi,\;c=\cos\theta\tag{1a,b,c}$$where ##0\leq\theta\leq\pi,0\leq\phi<2\pi\,##. The second constraint then becomes an equation interrelating ##\theta## and ##\phi\,##:$$1=a^{3}+b^{3}+c^{3}=\cos^{3}\theta+\sin^{3}\theta\left(\cos^{3}\phi+\sin^{3}\phi\right)\tag{2}$$the solution of which will allow us to complete the assigned task of finding:$$x\equiv a+b+c=\cos\theta+\sin\theta\left(\cos\phi+\sin\phi\right)\tag{3}$$An obvious solution of (2) is found by setting ##\text{sin}\theta=0\Rightarrow\theta=0## (since ##\text{cos}\pi## has the wrong sign) and ##\phi## arbitrary. Next, for ##\theta\neq 0## we can divide eq.(2) by ##\text{sin}^3\theta## to get:$$f\left(\theta\right)\equiv\csc^{3}\theta-\cot^{3}\theta=\sin^{3}\phi+\cos^{3}\phi\equiv g\left(\phi\right)\tag{4}$$To solve this, it's important to observe that, almost everywhere, ##f\left(\theta\right)>1## but ##g\left(\phi\right)<1##:
That means eq.(4) has only two isolated solution pairs ##\theta,\phi\,##, which I enumerate below, along with the ##\theta=0## case:\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \theta & \phi & f\left(\theta\right) & g\left(\phi\right) & a & b & c & x\\
\hline 0 & \text{arb.} & - & - & 0 & 0 & 1 & 1\\
\hline \frac{\pi}{2} & 0 & 1 & 1 & 1 & 0 & 0 & 1\\
\hline \frac{\pi}{2} & \frac{\pi}{2} & 1 & 1 & 0 & 1 & 0 & 1\\
\hline
\end{array}Evidently there are just three possible solutions for ##\left(a,b,c\right)##, namely: ##\left(1,0,0\right),\left(0,1,0\right),\left(0,0,1\right)##, all of which give ##x=a+b+c=1##.
OK, that's all folks!
(Edited on 2/6 to eliminate two spurious solutions arising from having erroneously doubled the range of ##\theta## (duh!).)