Finding number of real solutions The question is to  find number of real solutions of $F(x)=x^3+1=2\sqrt[3]{2x-1}=G(x)$
So first i tried to find the roots of it's derivative , which was not helpful as the equation formed was hard to solve. Then i tried to visualize the graph which was again not very helpful.
Luckily i found the above equation is true for $x=1$. At $x=0$, $G(x)$ is under $F(x)$ but it's derivative is greater than $F(x)$ till $x=1$ (easy to show). And at $x=1$ also derivative of $G(x)$ is greater than that of $F(x)$, so here $G(x)$ overtakes (not before $x=1$) $F(x)$.
Now after $x=1$, once the $F'(x)$ becomes greater than $G(x)$, it would remain like that so $F(x)$ once again over take $G(x)$ and hence we get 2 solutions of for $x>0$.
Then again when we go from $x=0$ to negative $x$-axis , we will see once $F(x)$ starts decreasing more rapidly than $G(x)$ it would remain that way, so we will get one solution for $x<0$.
So i got the correct answer of $3$. But this is a lucky solution. I would be happy if you can provide me with any type of solution. There should be a better way to solve this problem right?
 A: HINT.-Taking equivalently $\dfrac{x^3+1}{2}=\sqrt[3]{2x-1}$ it is verified that if you made $y=x$ you do have in both sides the same cubic equation; in fact:
$$\dfrac{x^3+1}{2}=x\iff x^3-2x+1=0\\\sqrt[3]{2x-1}=x\iff x^3-2x+1=0$$ so the two curves have its common points with the diagonal $y=x$. 
Consequently because it is clear that $(1,1)$ is one of this points the two others are the roots of the quadratic equation $x^2+x-1=0$.
Thus the roots are $x=1$ and $x=\dfrac{-1+\sqrt5}{2}$ 
A: your equation is :
$$\left(x^{3}+1\right)^{3}=8\left(2x-1\right)$$
Where:$$\left(x^{3}+1\right)^{3}=\left(x^{6}+2x^{3}+1\right)\left(x^{3}+1\right)=x^{9}+x^{6}+2x^{6}+2x^{3}+x^{3}+1=x^{9}+3x^{6}+3x^{3}+1$$
So we have:
$$x^{9}+3x^{6}+3x^{3}+1=16x-8$$
$$x^{9}+3x^{6}+3x^{3}-16x+9=0$$
Assume rational roots of this equations are in the form $\frac{p}{q}$ where $p,q∈ℤ$ and $q≠0$, also assume this fraction is in the simplest form (GCD of $p,q$ is $1$), using rational root theorem implies $p$ must divide $9$ and $q$ must divide $1$, so the whole fractions with these assumptions are:
$$\pm1 , \pm3 ,\pm9$$
Checking them implies $\color{green} {\boxed {x=1}}$ is one of the solutions.
Now you can divide the equation by $x-3$ to get :
$x^{8}+x^{7}+x^{6}+4x^{5}+4x^{4}+4x^{3}+7x^{2}+7x-9$
Also for solving $x^{8}+x^{7}+x^{6}+4x^{5}+4x^{4}+4x^{3}+7x^{2}+7x-9$ I suggest to use Newton's method, then you can get all the solutions.
For more numericals algorithms (but more difficult ways) you can use  , Bisection method , Muller's method,Secant method,false position method
