The length of the latus rectum of the curve represented by $x = 3(\cos t + \sin t)$ and $y = 4(\cos t - \sin t)$ is:

Given parametric equations are:

$x = 3(\cos t + \sin t) \Rightarrow \dfrac{x}{3} = \cos t + \sin t$

$y = 4(\cos t - \sin t) \Rightarrow \dfrac{y}{4} = \cos t - \sin t$

Squaring and adding both equations, we get:

$\left(\dfrac{x}{3}\right)^2 + \left(\dfrac{y}{4}\right)^2 = (\cos t + \sin t)^2 + (\cos t - \sin t)^2$

$\dfrac{x^2}{9} + \dfrac{y^2}{16} = 2(\cos^2 t + \sin^2 t)$

$\dfrac{x^2}{9} + \dfrac{y^2}{16} = 2$

$\dfrac{x^2}{18} + \dfrac{y^2}{32} = 1$

This represents an ellipse with $a^2 = 18$ and $b^2 = 32$.

Since $b^2 > a^2$, the major axis lies along the y-axis.

The length of the latus rectum is given by $\dfrac{2a^2}{b}$.

Substituting the values:

Length of latus rectum $= \dfrac{2 \times 18}{\sqrt{32}} = \dfrac{36}{4\sqrt{2}} = \dfrac{9}{\sqrt{2}}$

Answer: $\dfrac{9}{\sqrt{2}}$