1. **Problem Statement:** We are given positive real numbers $x, y, z$ such that: $$x + y + z = 12$$ $$x^2 + y^2 + z^2 = 56$$ We need to find the maximum possible value of: $$P = xyz$$ 2. **Relevant Formulas and Rules:** - The Arithmetic Mean - Geometric Mean (AM-GM) inequality states that for positive real numbers: $$\frac{x + y + z}{3} \geq \sqrt[3]{xyz}$$ - Equality holds when $x = y = z$. - We also use the identity: $$ (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) $$ 3. **Step 1: Find $xy + yz + zx$** Given: $$ (x + y + z)^2 = 12^2 = 144 $$ Using the identity: $$ 144 = 56 + 2(xy + yz + zx) $$ Rearranging: $$ 2(xy + yz + zx) = 144 - 56 = 88 $$ $$ xy + yz + zx = 44 $$ 4. **Step 2: Use symmetric sums and consider $x, y, z$ as roots of a cubic** Let $x, y, z$ be roots of: $$ t^3 - (x+y+z)t^2 + (xy + yz + zx)t - xyz = 0 $$ So the cubic is: $$ t^3 - 12t^2 + 44t - P = 0 $$ 5. **Step 3: Use inequality to find maximum $P$** By AM-GM: $$ \frac{12}{3} = 4 \geq \sqrt[3]{P} $$ Cubing both sides: $$ 64 \geq P $$ So $P \leq 64$. 6. **Step 4: Check if $P=64$ is possible given the constraints** If $x = y = z = 4$, then: $$ x^2 + y^2 + z^2 = 3 \times 4^2 = 48 \neq 56 $$ So equal values do not satisfy the second condition. 7. **Step 5: Use Lagrange multipliers or consider $x = y$ to simplify** Assume $x = y = a$, then $z = 12 - 2a$. Given: $$ 2a^2 + (12 - 2a)^2 = 56 $$ Expanding: $$ 2a^2 + 144 - 48a + 4a^2 = 56 $$ $$ 6a^2 - 48a + 144 = 56 $$ $$ 6a^2 - 48a + 88 = 0 $$ Divide by 2: $$ 3a^2 - 24a + 44 = 0 $$ 8. **Step 6: Solve quadratic for $a$** Discriminant: $$ \Delta = (-24)^2 - 4 \times 3 \times 44 = 576 - 528 = 48 $$ $$ a = \frac{24 \pm \sqrt{48}}{2 \times 3} = \frac{24 \pm 4\sqrt{3}}{6} = 4 \pm \frac{2\sqrt{3}}{3} $$ 9. **Step 7: Calculate $P = a^2 (12 - 2a)$ for both roots** For $a = 4 - \frac{2\sqrt{3}}{3}$: $$ z = 12 - 2a = 12 - 2\left(4 - \frac{2\sqrt{3}}{3}\right) = 12 - 8 + \frac{4\sqrt{3}}{3} = 4 + \frac{4\sqrt{3}}{3} $$ Calculate $P$: $$ P = a^2 z = \left(4 - \frac{2\sqrt{3}}{3}\right)^2 \left(4 + \frac{4\sqrt{3}}{3}\right) $$ For $a = 4 + \frac{2\sqrt{3}}{3}$, $z$ becomes negative, which is invalid since $z$ must be positive. 10. **Step 8: Simplify $P$ for valid $a$** Calculate $a^2$: $$ a^2 = \left(4 - \frac{2\sqrt{3}}{3}\right)^2 = 16 - 2 \times 4 \times \frac{2\sqrt{3}}{3} + \left(\frac{2\sqrt{3}}{3}\right)^2 = 16 - \frac{16\sqrt{3}}{3} + \frac{4 \times 3}{9} = 16 - \frac{16\sqrt{3}}{3} + \frac{4}{3} = \frac{48}{3} - \frac{16\sqrt{3}}{3} + \frac{4}{3} = \frac{52 - 16\sqrt{3}}{3} $$ Calculate $z$: $$ z = 4 + \frac{4\sqrt{3}}{3} = \frac{12 + 4\sqrt{3}}{3} $$ So: $$ P = a^2 z = \frac{52 - 16\sqrt{3}}{3} \times \frac{12 + 4\sqrt{3}}{3} = \frac{(52 - 16\sqrt{3})(12 + 4\sqrt{3})}{9} $$ Multiply numerator: $$ 52 \times 12 = 624 $$ $$ 52 \times 4\sqrt{3} = 208\sqrt{3} $$ $$ -16\sqrt{3} \times 12 = -192\sqrt{3} $$ $$ -16\sqrt{3} \times 4\sqrt{3} = -16 \times 4 \times 3 = -192 $$ Sum: $$ 624 + 208\sqrt{3} - 192\sqrt{3} - 192 = (624 - 192) + (208\sqrt{3} - 192\sqrt{3}) = 432 + 16\sqrt{3} $$ Therefore: $$ P = \frac{432 + 16\sqrt{3}}{9} = 48 + \frac{16\sqrt{3}}{9} $$ 11. **Final answer:** $$ \boxed{P_{max} = 48 + \frac{16\sqrt{3}}{9}} $$ This is the maximum possible value of $xyz$ under the given constraints.