1. The problem asks to evaluate the definite integral $$\int_{-a}^a (x^3 + x) \, dx$$. 2. Recall that the integral of a sum is the sum of the integrals: $$\int_{-a}^a (x^3 + x) \, dx = \int_{-a}^a x^3 \, dx + \int_{-a}^a x \, dx$$. 3. Notice that both $x^3$ and $x$ are odd functions because $f(-x) = -f(x)$ for these functions. 4. The integral of an odd function over symmetric limits $[-a, a]$ is zero: $$\int_{-a}^a f(x) \, dx = 0 \quad \text{if } f \text{ is odd}$$. 5. Since both $x^3$ and $x$ are odd, their integrals over $[-a, a]$ are zero: $$\int_{-a}^a x^3 \, dx = 0$$ $$\int_{-a}^a x \, dx = 0$$. 6. Therefore, the value of the integral is: $$\int_{-a}^a (x^3 + x) \, dx = 0 + 0 = 0$$. Final answer: $$0$$.