Question: Solve the polynomial equation
16y^3 - 9 = y - 144y^2
1. **State the problem:** Solve the polynomial equation $$16y^3 - 9 = y - 144y^2$$.
2. **Rewrite the equation:** Move all terms to one side to set the equation to zero:
$$16y^3 - 9 - y + 144y^2 = 0$$
3. **Combine like terms:**
$$16y^3 + 144y^2 - y - 9 = 0$$
4. **Try to factor by grouping:** Group terms:
$$(16y^3 + 144y^2) + (-y - 9) = 0$$
Factor out common factors in each group:
$$16y^2(y + 9) - 1(y + 9) = 0$$
5. **Factor out the common binomial:**
$$(16y^2 - 1)(y + 9) = 0$$
6. **Set each factor equal to zero:**
$$16y^2 - 1 = 0$$
$$y + 9 = 0$$
7. **Solve each equation:**
For $$16y^2 - 1 = 0$$:
$$16y^2 = 1$$
$$y^2 = \frac{1}{16}$$
$$y = \pm \frac{1}{4}$$
For $$y + 9 = 0$$:
$$y = -9$$
8. **Final solutions:**
$$y = \frac{1}{4}, \quad y = -\frac{1}{4}, \quad y = -9$$