1. The problem is to find the $x$-coordinates where the gradient (derivative) of the curve $y=4x^3 - 8x + 5$ equals $\frac{1}{3}$. 2. First, find the derivative of $y$ with respect to $x$, which represents the gradient: $$\frac{dy}{dx} = 12x^2 - 8$$ 3. Set the derivative equal to $\frac{1}{3}$ to find the points where the gradient is $\frac{1}{3}$: $$12x^2 - 8 = \frac{1}{3}$$ 4. Solve for $x$ by multiplying both sides by 3 to clear the fraction: $$3(12x^2 - 8) = 1 \implies 36x^2 - 24 = 1$$ 5. Add 24 to both sides: $$36x^2 = 25$$ 6. Divide both sides by 36: $$x^2 = \frac{25}{36}$$ 7. Take the square root of both sides: $$x = \pm \frac{5}{6}$$ 8. Therefore, the two $x$-coordinates where the gradient is $\frac{1}{3}$ are: $$x = \frac{5}{6} \text{ and } x = -\frac{5}{6}$$