Let the vector make angles $\alpha$, $\beta$, $\gamma$ with the positive directions of axes OX, OY and OZ respectively. Since the vector is equally inclined to all three axes, we have $\alpha = \beta = \gamma$.

Let $\alpha = \beta = \gamma = \theta$.

The direction cosines of a vector are $\cos\alpha$, $\cos\beta$ and $\cos\gamma$. We know that for any vector, the sum of squares of direction cosines equals unity:

$$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$$

Substituting $\alpha = \beta = \gamma = \theta$:

$$\cos^2\theta + \cos^2\theta + \cos^2\theta = 1$$ $$3\cos^2\theta = 1$$ $$\cos^2\theta = \dfrac{1}{3}$$ $$\cos\theta = \pm\dfrac{1}{\sqrt{3}}$$

Therefore, the direction cosines are:

$$\cos\alpha = \cos\beta = \cos\gamma = \pm\dfrac{1}{\sqrt{3}}$$

Hence, the direction cosines of a vector equally inclined to the axes OX, OY and OZ are $\pm\left(\dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}\right)$.