Verifying that $u(x, t) = \frac{1}{2}c \text{sech}^2\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)$ is a solution to $u_t + 6uu_x + u_{xxx} = 0$ I'm trying to verify that 
$$u(x, t) = \frac{1}{2}c \text{sech}^2\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)$$
is a solution to the KdV equation
$$u_t + 6uu_x + u_{xxx} = 0$$
My calculations have lead to 
$$u_t = \frac{c^{5/2}}{2}\text{sech}^2\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)\text{tanh}\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)$$
$$u_x = -\frac{c^{3/2}}{2}\text{sech}^2\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)\text{tanh}\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)$$
$$u_{xxx} = -\frac{c^{5/2}}{2}\text{sech}^2\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)\text{tanh}^3\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right) + c^{5/2}\text{sech}^4\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)\text{tanh}\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)$$
I have used Symbolab to verify that these are correct.
So we have 
$$u_t + 6uu_x + u_{xxx} = \frac{c^{5/2}}{2}\text{sech}^2\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)\text{tanh}\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right) + 6 \left[ \frac{1}{2}c \text{sech}^2\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right) \right] \left[ -\frac{c^{3/2}}{2}\text{sech}^2\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)\text{tanh}\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right) \right] + \left[ \left( -\frac{c^{5/2}}{2}\text{sech}^2\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)\text{tanh}^3\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right) + c^{5/2}\text{sech}^4\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right)\text{tanh}\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right) \right) \right]$$ 
Which, as far as I can tell, does not equal $0$?
I'm wondering if there's an error that I'm not noticing, or whether there is some hyperbolic trig identity that I'm supposed to use?

Begin by setting $a=\text{sech}^2\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right) $and $b=\text{tanh}\left(\frac{1}{2} \sqrt{c} (x - ct - x_0)\right).$ If $c=0,$ then the function in question is clearly a solution, hence we may factor out $c^{5/2}/2.$
The computed expression simplifies to $ab-3a^2b-ab^3+2a^2b = ab-a^2b-ab^3 = ab(1-a-b^2) = 0,$ where we used the trig identity $a+b^2=1.$
