To solve this problem, we need to compute $q_3$, the next orthogonal polynomial in the sequence, and then express $x^3$ as a linear combination of the polynomials $q_0, q_1, q_2, q_3$.

Step 1: Orthogonality Condition

The polynomials $\{q_i\}$ are orthogonal with respect to the standard square norm on the interval $[-1, 1]$, which means that for any two different polynomials $q_i$ and $q_j$:

$\int_{-1}^{1} q_i(x) q_j(x) \, dx = 0 \quad \text{for } i \neq j$

Step 2: Determine $q_3(x)$

Assume that $q_3(x)$ has the general form of a cubic polynomial:

$q_3(x) = x^3 + a_2 x^2 + a_1 x + a_0$

The orthogonality condition requires $q_3(x)$ to be orthogonal to $q_0(x) = 1$, $q_1(x) = x$, and $q_2(x) = x^2 - \frac{1}{3}$.

Let's calculate the coefficients $a_2$, $a_1$, and $a_0$ by ensuring orthogonality with respect to $q_0(x)$, $q_1(x)$, and $q_2(x)$.

Orthogonality with $q_0(x) = 1$:

$\int_{-1}^{1} q_3(x) \cdot 1 \, dx = \int_{-1}^{1} \left( x^3 + a_2 x^2 + a_1 x + a_0 \right) dx = 0$

Evaluating the integrals:

$\int_{-1}^{1} x^3 \, dx = 0, \quad \int_{-1}^{1} x^2 \, dx = \frac{2}{3}, \quad \int_{-1}^{1} x \, dx = 0, \quad \int_{-1}^{1} 1 \, dx = 2$

This gives:

$a_2 \cdot \frac{2}{3} + 2a_0 = 0 \quad \Rightarrow \quad a_0 = -\frac{a_2}{3}$

Orthogonality with $q_1(x) = x$:

$\int_{-1}^{1} q_3(x) \cdot x \, dx = \int_{-1}^{1} \left( x^4 + a_2 x^3 + a_1 x^2 + a_0 x \right) dx = 0$

Evaluating the integrals:

$\int_{-1}^{1} x^4 \, dx = \frac{2}{5}, \quad \int_{-1}^{1} x^3 \, dx = 0, \quad \int_{-1}^{1} x^2 \, dx = \frac{2}{3}, \quad \int_{-1}^{1} x \, dx = 0$

This simplifies to:

$a_1 \cdot \frac{2}{3} = 0 \quad \Rightarrow \quad a_1 = 0$

Orthogonality with $q_2(x) = x^2 - \frac{1}{3}$:

[ \int_{-1}^{1} q_3(x) \cdot \left(x^2 - \frac{1}{3}\right) dx = \int_{-1