Step 1: Apply binomial theorem to expand both expressions.

($\sqrt{3}$ + $\sqrt{2}$)⁶ = ⁶C₀($\sqrt{3}$)⁶ + ⁶C₁($\sqrt{3}$)⁵($\sqrt{2}$) + ⁶C₂($\sqrt{3}$)⁴($\sqrt{2}$)² + ⁶C₃($\sqrt{3}$)³($\sqrt{2}$)³ + ⁶C₄($\sqrt{3}$)²($\sqrt{2}$)⁴ + ⁶C₅($\sqrt{3}$)($\sqrt{2}$)⁵ + ⁶C₆($\sqrt{2}$)⁶

($\sqrt{3}$ - $\sqrt{2}$)⁶ = ⁶C₀($\sqrt{3}$)⁶ - ⁶C₁($\sqrt{3}$)⁵($\sqrt{2}$) + ⁶C₂($\sqrt{3}$)⁴($\sqrt{2}$)² - ⁶C₃($\sqrt{3}$)³($\sqrt{2}$)³ + ⁶C₄($\sqrt{3}$)²($\sqrt{2}$)⁴ - ⁶C₅($\sqrt{3}$)($\sqrt{2}$)⁵ + ⁶C₆($\sqrt{2}$)⁶

Step 2: Subtract the two expansions.

When subtracting, terms with even powers of $\sqrt{2}$ cancel out, and terms with odd powers of $\sqrt{2}$ double.

($\sqrt{3}$ + $\sqrt{2}$)⁶ - ($\sqrt{3}$ - $\sqrt{2}$)⁶ = 2[⁶C₁($\sqrt{3}$)⁵($\sqrt{2}$) + ⁶C₃($\sqrt{3}$)³($\sqrt{2}$)³ + ⁶C₅($\sqrt{3}$)($\sqrt{2}$)⁵]

Step 3: Calculate the binomial coefficients and simplify each term.

⁶C₁ = 6, ⁶C₃ = 20, ⁶C₅ = 6

⁶C₁($\sqrt{3}$)⁵($\sqrt{2}$) = $$6 \times 9$$$\sqrt{3}$×$\sqrt{2}$= 54$\sqrt{6}$

⁶C₃($\sqrt{3}$)³($\sqrt{2}$)³ = $20 \times 3$√$$3 \times 2$$$\sqrt{2}$= 120$\sqrt{6}$

⁶C₅($\sqrt{3}$)($\sqrt{2}$)⁵ = 6 × √$$3 \times 4$$$\sqrt{2}$= 24$\sqrt{6}$

Step 4: Add all terms and multiply by 2.

= 2[54$\sqrt{6}$ + 120$\sqrt{6}$ + 24$\sqrt{6}$]

= 2[198$\sqrt{6}$]

= 396$\sqrt{6}$