1. **State the problem:** We need to find the values of $x$ that must be excluded from the domain of the expression $$\frac{x}{3x^3 + 77}$$ because division by zero is undefined. 2. **Identify the restriction:** The denominator cannot be zero. So, solve the equation: $$3x^3 + 77 = 0$$ 3. **Solve for $x$:** $$3x^3 = -77$$ $$x^3 = -\frac{77}{3}$$ $$x = \sqrt[3]{-\frac{77}{3}}$$ 4. **Evaluate the cube root:** The cube root of a negative number is negative, so $$x = -\sqrt[3]{\frac{77}{3}}$$ 5. **Check the given options:** The options are 0, 3, and -1. - For $x=0$, denominator is $3(0)^3 + 77 = 77 \neq 0$. - For $x=3$, denominator is $3(3)^3 + 77 = 3(27) + 77 = 81 + 77 = 158 \neq 0$. - For $x=-1$, denominator is $3(-1)^3 + 77 = 3(-1) + 77 = -3 + 77 = 74 \neq 0$. None of these values make the denominator zero. 6. **Conclusion:** The value that must be excluded is $x = -\sqrt[3]{\frac{77}{3}}$, which is not among the options given. **Final answer:** None of the values 0, 3, or -1 need to be excluded from the domain.