The problem involves finding the next orthogonal polynomial $q_3(x)$ given the first three orthogonal polynomials $q_0(x) = 1$, $q_1(x) = x$, and $q_2(x) = x^2 - \frac{1}{3}$. We will use the Gram-Schmidt process to ensure the orthogonality of the polynomials over the interval $[-1, 1]$ with respect to the standard square norm:

$\langle f, g \rangle = \int_{-1}^{1} f(x) g(x) \, dx$

Step 1: Find $q_3(x)$

We assume $q_3(x)$ is a linear combination of monomials up to $x^3$ because the degree of $q_3(x)$ should be 3. Hence, let:

$q_3(x) = x^3 + A x^2 + B x + C$

To ensure orthogonality, $q_3(x)$ must satisfy:

$\langle q_3, q_0 \rangle = 0, \quad \langle q_3, q_1 \rangle = 0, \quad \langle q_3, q_2 \rangle = 0$

Orthogonality with $q_0(x)$:

$\langle q_3, q_0 \rangle = \int_{-1}^{1} (x^3 + A x^2 + B x + C) \cdot 1 \, dx = \int_{-1}^{1} (x^3 + A x^2 + B x + C) \, dx$

[ = \left[ \frac{x^4}{4} \right]{-1}^{1} + A \left[ \frac{x^3}{3} \right]{-1}^{1} + B \left[ \frac{x^2}{2} \right]{-1}^{1} + C \left[ x \right]{-1}^{1} ]

Since all odd powers of $x$ vanish when integrated over $[-1, 1]$, we get:

$\langle q_3, q_0 \rangle = 0 + 0 + 2B \cdot \frac{1}{2} + 0 = B$

Thus, $B = 0$.

Orthogonality with $q_1(x)$:

$\langle q_3, q_1 \rangle = \int_{-1}^{1} (x^3 + A x^2 + C) \cdot x \, dx = \int_{-1}^{1} (x^4 + A x^3 + C x) \, dx$

[ = \left[ \frac{x^5}{5} \right]{-1}^{1} + A \left[ \frac{x^4}{4} \right]{-1}^{1} + C \left[ \frac{x^2}{2} \right]_{-1}^{1} ]

Again, odd powers vanish, and we are left with:

$\langle q_3, q_1 \rangle = 0 + A \cdot \frac{2}{4} + 0 = \frac{A}{2}$

Thus, $A = 0$.

Orthogonality with $q_2(x)$:

$q_2(x) = x^2 - \frac{1}{3}$ $\langle q_3, q_2 \rangle = \int_{-1}^{1} (x^3 + C) \cdot (x^2 - \frac{1}{3}) \, dx = \int_{-1}^{1} (x^5 - \frac{1}{3} x^3 + Cx^2 - \frac{C}{3}) \, dx$

This simplifies to:

$\langle q_3, q_2 \rangle = 0 + \frac{C}{3} \cdot 2 - 0 = \frac{2C}{3}$

Thus, $C = 0$.

Conclusion for $q_3(x)$:

Therefore, $q_3(x) = x^3$, but we need to normalize it to ensure it satisfies the orthogonality conditions correctly. Let’s calculate the norm of $x^3$:

$\| q_3 \|^2 = \langle x^3, x^3 \rangle = \int_{-1}^{1} x^6 \, dx = \left[ \frac{x^7}{7} \right]_{-1}^{1} = \frac{2}{7}$

Thus, the normalized polynomial is:

$q_3(x) = \sqrt{\frac{7}{2}} x^3$

Step 2: Express $x^3$ as a linear combination of $q_0, q_1, q_2, q_3$

We can now express $x^3$ as a linear combination of the orthogonal polynomials:

$x^3 = c_0 q_0 + c_1 q_1 + c_2 q_2 + c_3 q_3$

Where $q_0 = 1$, $q_1 = x$, $q_2 = x^2 - \frac{1}{3}$, and $q_3 = \sqrt{\frac{7}{2}} x^3$.

We now need to compute the coefficients $c_0, c_1, c_2, c_3$ by projecting $x^3$ onto each of the polynomials.