If $\tan \mathrm{A}=\frac{1}{\sqrt{x\left(x^2+x+1\right)}}, \tan \mathrm{B}=\frac{\sqrt{x}}{\sqrt{x^2+x+1}}$ and $\tan \mathrm{C}=\sqrt{x^{-1}+x^{-2}+x^{-3}}$ then

Simplify $\tan C$:
Given $\tan C = \sqrt{x^{-1}+x^{-2}+x^{-3}}$, express with common denominator $x^3$:

$\tan C = \sqrt{\frac{x^2+x+1}{x^3}} = \frac{\sqrt{x^2+x+1}}{x^{3/2}}$

Compute $\tan(A+B)$:
Using $\tan A = \frac{1}{\sqrt{x}\sqrt{x^2+x+1}}$ and $\tan B = \frac{\sqrt{x}}{\sqrt{x^2+x+1}}$, the sum is:

$\tan A + \tan B = \frac{1+x}{\sqrt{x}\sqrt{x^2+x+1}}$

The product simplifies to:

$\tan A \tan B = \frac{1}{x^2+x+1}$

Applying the tangent addition formula:

$\tan(A+B) = \frac{\frac{1+x}{\sqrt{x}\sqrt{x^2+x+1}}}{1 - \frac{1}{x^2+x+1}} = \frac{1+x}{\sqrt{x}\sqrt{x^2+x+1}} \cdot \frac{x^2+x+1}{x^2+x}$

Factoring and simplifying:

$\tan(A+B) = \frac{1+x}{\sqrt{x}\sqrt{x^2+x+1}} \cdot \frac{(\sqrt{x^2+x+1})^2}{x(1+x)} = \frac{\sqrt{x^2+x+1}}{x^{3/2}}$

Conclusion:
Since $\tan(A+B) = \tan C$, it follows that $A+B=C$.