Calculation of $x$ in $\tan^{-1}\left(\frac{x}{1-x^2}\right)+\tan^{-1}\left(\frac{1}{x^3}\right) = \frac{3\pi}{4}$ The no. of real values of $x$ satisfying $\displaystyle \tan^{-1}\left(\frac{x}{1-x^2}\right)+\tan^{-1}\left(\frac{1}{x^3}\right) = \frac{3\pi}{4}$
options :: 
(a) $0$
(b) $1$
(c) $2$
(d) Infinitely many
My Try:: Using The formula $\tan^{-1}A+\tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$
So $\displaystyle \tan^{-1}\left(\frac{\frac{x}{1-x^2}+\frac{1}{x^3}}{1-\frac{x}{x^3.(1-x^2)}}\right) = \frac{3\pi}{4}$
So $\displaystyle \frac{x^4+1-x^2}{x^3-x^5-x}= -1$
So $(x-1)(x^4-x+1) = 0$ 
$x = 1$ and $x^4 -x+1 = 0$
So Iam getting only one value of $x$ which is $x=1$ and How can i calculate any real value of $x$ exists from $x^4-x+1 = 0$
OR any other method by using we can solve this Question
Thanks          

*

*If $|x|<1$, then $x^4+1>1>x$. If $|x|\geq 1$, then $x^4+1>x^4\geq x$. In both cases 
$$
x^4-x+1>0
$$
so the equation $x^4-x+1=0$ have no real solutions.

*Note that $x=1$ is not a solution of original equation.
