1. **State the problem:** Solve the equation $x^3 - x^2 = 80$ for $x$. 2. **Rewrite the equation:** Move all terms to one side to set the equation to zero: $$x^3 - x^2 - 80 = 0$$ 3. **Look for rational roots:** Use the Rational Root Theorem, possible roots are factors of 80: $\pm1, \pm2, \pm4, \pm5, \pm8, \pm10, \pm16, \pm20, \pm40, \pm80$. 4. **Test possible roots:** Try $x=5$: $$5^3 - 5^2 - 80 = 125 - 25 - 80 = 20 \neq 0$$ Try $x=4$: $$4^3 - 4^2 - 80 = 64 - 16 - 80 = -32 \neq 0$$ Try $x= -4$: $$(-4)^3 - (-4)^2 - 80 = -64 - 16 - 80 = -160 \neq 0$$ Try $x= 8$: $$8^3 - 8^2 - 80 = 512 - 64 - 80 = 368 \neq 0$$ Try $x= 10$: $$10^3 - 10^2 - 80 = 1000 - 100 - 80 = 820 \neq 0$$ Try $x= -5$: $$(-5)^3 - (-5)^2 - 80 = -125 - 25 - 80 = -230 \neq 0$$ Try $x= 1$: $$1 - 1 - 80 = -80 \neq 0$$ Try $x= -1$: $$-1 - 1 - 80 = -82 \neq 0$$ Try $x= 2$: $$8 - 4 - 80 = -76 \neq 0$$ Try $x= -2$: $$-8 - 4 - 80 = -92 \neq 0$$ Try $x= 0$: $$0 - 0 - 80 = -80 \neq 0$$ 5. **Use numerical methods or graphing:** Since no rational root found, approximate solution numerically. 6. **Approximate root:** By testing values between 4 and 5, since at 4 the value is negative and at 5 positive, root lies between 4 and 5. 7. **Final answer:** The real root of $x^3 - x^2 = 80$ is approximately $x \approx 4.7$ (rounded to one decimal place).