1. **State the problem:** Solve the expression $x^3 - 5$ using first principles. 2. **Understand first principles approach:** This typically means evaluating or simplifying directly from definitions or basic operations rather than applying shortcuts. 3. Since $x^3 - 5$ is a simple algebraic expression, to find its value, substitute any value for $x$ and simplify. 4. For example, if $x=2$, then $2^3 - 5 = 8 - 5 = 3$. 5. If the problem meant to find the derivative of $f(x) = x^3 - 5$ using first principles, proceed as follows: 6. Use the definition of derivative: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 7. Substitute $f(x) = x^3 - 5$: $$f'(x) = \lim_{h \to 0} \frac{(x+h)^3 - 5 - (x^3 - 5)}{h} = \lim_{h \to 0} \frac{(x+h)^3 - x^3}{h}$$ 8. Expand $(x+h)^3$: $$(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$$ 9. Substitute back: $$f'(x) = \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h} = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h}$$ 10. Factor out $h$: $$= \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2)}{h} = \lim_{h \to 0} (3x^2 + 3xh + h^2)$$ 11. Taking the limit as $h \to 0$: $$f'(x) = 3x^2$$ 12. **Final answer:** The derivative of $x^3 - 5$ using first principles is $3x^2$.