1. **Problem 1: Solve $x^3 - 64 = 0$** This is a difference of cubes since $64 = 4^3$. The formula for difference of cubes is: $$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$ Here, $a = x$ and $b = 4$. 2. Apply the formula: $$x^3 - 4^3 = (x - 4)(x^2 + 4x + 16) = 0$$ 3. Set each factor equal to zero: - $x - 4 = 0 \implies x = 4$ - $x^2 + 4x + 16 = 0$ 4. Solve the quadratic $x^2 + 4x + 16 = 0$ using the discriminant: $$\Delta = b^2 - 4ac = 4^2 - 4 \times 1 \times 16 = 16 - 64 = -48 < 0$$ Since the discriminant is negative, no real roots here. **Final real solution:** $x = 4$ 1. **Problem 2: Solve $x^3 - 27 = 0$** Again, difference of cubes with $27 = 3^3$. 2. Use formula: $$x^3 - 3^3 = (x - 3)(x^2 + 3x + 9) = 0$$ 3. Set factors to zero: - $x - 3 = 0 \implies x = 3$ - $x^2 + 3x + 9 = 0$ 4. Solve quadratic: $$\Delta = 3^2 - 4 \times 1 \times 9 = 9 - 36 = -27 < 0$$ No real roots here. **Final real solution:** $x = 3$ 1. **Problem 3: Solve $x^4 - 8x^2 = -16$** Rewrite: $$x^4 - 8x^2 + 16 = 0$$ 2. Let $y = x^2$, then: $$y^2 - 8y + 16 = 0$$ 3. Factor or use quadratic formula: $$\Delta = (-8)^2 - 4 \times 1 \times 16 = 64 - 64 = 0$$ 4. One repeated root: $$y = \frac{8}{2} = 4$$ 5. Recall $y = x^2$, so: $$x^2 = 4 \implies x = \pm 2$$ **Final solutions:** $x = 2, -2$ 1. **Problem 4: Solve $x^4 - 12x^2 = 64$** Rewrite: $$x^4 - 12x^2 - 64 = 0$$ 2. Let $y = x^2$, then: $$y^2 - 12y - 64 = 0$$ 3. Use quadratic formula: $$y = \frac{12 \pm \sqrt{(-12)^2 - 4 \times 1 \times (-64)}}{2} = \frac{12 \pm \sqrt{144 + 256}}{2} = \frac{12 \pm \sqrt{400}}{2}$$ 4. Simplify: $$y = \frac{12 \pm 20}{2}$$ 5. Two roots: - $y = \frac{12 + 20}{2} = 16$ - $y = \frac{12 - 20}{2} = -4$ 6. Since $y = x^2$, discard negative root $y = -4$ (no real $x$). 7. Solve $x^2 = 16$: $$x = \pm 4$$ **Final solutions:** $x = 4, -4$ 1. **Problem 5: Solve $x^3 + 3x^2 + 4x + 12 = 0$** Try factoring by grouping: $$(x^3 + 3x^2) + (4x + 12) = x^2(x + 3) + 4(x + 3) = (x^2 + 4)(x + 3) = 0$$ Set each factor to zero: - $x + 3 = 0 \implies x = -3$ - $x^2 + 4 = 0 \implies x^2 = -4$ no real roots **Final real solution:** $x = -3$ 1. **Problem 6: Simplify $6x^2 y + 8x^2 + 3xy + 4x$** Group terms: $$(6x^2 y + 3xy) + (8x^2 + 4x) = 3xy(2x + 1) + 4x(2x + 1) = (3xy + 4x)(2x + 1)$$ Factor out $x$ in first bracket: $$x(3y + 4)(2x + 1)$$ **Final factorization:** $x(3y + 4)(2x + 1)$ 1. **Problem 7: Solve $x^3 + 13x = 10x^2$** Rewrite: $$x^3 - 10x^2 + 13x = 0$$ Factor out $x$: $$x(x^2 - 10x + 13) = 0$$ Set factors to zero: - $x = 0$ - Solve quadratic $x^2 - 10x + 13 = 0$ Discriminant: $$\Delta = (-10)^2 - 4 \times 1 \times 13 = 100 - 52 = 48$$ Roots: $$x = \frac{10 \pm \sqrt{48}}{2} = \frac{10 \pm 4\sqrt{3}}{2} = 5 \pm 2\sqrt{3}$$ **Final solutions:** $x = 0, 5 + 2\sqrt{3}, 5 - 2\sqrt{3}$ 1. **Problem 8: Solve $x^3 - 6x^2 + 6x = 0$** Factor out $x$: $$x(x^2 - 6x + 6) = 0$$ Set factors to zero: - $x = 0$ - Solve quadratic $x^2 - 6x + 6 = 0$ Discriminant: $$\Delta = (-6)^2 - 4 \times 1 \times 6 = 36 - 24 = 12$$ Roots: $$x = \frac{6 \pm \sqrt{12}}{2} = \frac{6 \pm 2\sqrt{3}}{2} = 3 \pm \sqrt{3}$$ **Final solutions:** $x = 0, 3 + \sqrt{3}, 3 - \sqrt{3}$ 1. **Problem 9: Solve $12x^3 = 60x^2 + 75x$** Rewrite: $$12x^3 - 60x^2 - 75x = 0$$ Factor out $3x$: $$3x(4x^2 - 20x - 25) = 0$$ Set factors to zero: - $3x = 0 \implies x = 0$ - Solve quadratic $4x^2 - 20x - 25 = 0$ Discriminant: $$\Delta = (-20)^2 - 4 \times 4 \times (-25) = 400 + 400 = 800$$ Roots: $$x = \frac{20 \pm \sqrt{800}}{2 \times 4} = \frac{20 \pm 20\sqrt{2}}{8} = \frac{20}{8} \pm \frac{20\sqrt{2}}{8} = 2.5 \pm 2.5\sqrt{2}$$ **Final solutions:** $x = 0, 2.5 + 2.5\sqrt{2}, 2.5 - 2.5\sqrt{2}$ 1. **Problem 10: Solve $5x^3 = 5x^2 + 12x$** Rewrite: $$5x^3 - 5x^2 - 12x = 0$$ Factor out $x$: $$x(5x^2 - 5x - 12) = 0$$ Set factors to zero: - $x = 0$ - Solve quadratic $5x^2 - 5x - 12 = 0$ Discriminant: $$\Delta = (-5)^2 - 4 \times 5 \times (-12) = 25 + 240 = 265$$ Roots: $$x = \frac{5 \pm \sqrt{265}}{2 \times 5} = \frac{5 \pm \sqrt{265}}{10}$$ **Final solutions:** $x = 0, \frac{5 + \sqrt{265}}{10}, \frac{5 - \sqrt{265}}{10}$