To prove:

Formula used: (i)

(ii) (a+b)ⁿ =   ⁿC₀aⁿ + ⁿC₁a^n-1b + ⁿC₂a^n-2b² + …… +ⁿC_n-1ab^n-1 + ⁿC_nbⁿ

(a+b)⁴ = ⁴C₀a⁴ + ⁴C₁a^4-1b + ⁴C₂a^4-2b² + ⁴C₃a^4-3b³ + ⁴C₄b⁴

⇒ ⁴C₀a⁴ + ⁴C₁a³b + ⁴C₂a²b² + ⁴C₃a¹b³ + ⁴C₄b⁴ … (i)

(a-b)⁴ = ⁴C₀a⁴ + ⁴C₁a^4-1(-b) + ⁴C₂a^4-2(-b)² +⁴C₃a^4-3(-b)³+⁴C₄(-b)⁴

⇒ ⁴C₀a⁴ - ⁴C₁a³b + ⁴C₂a²b² - ⁴C₃ab³ + ⁴C₄b⁴ … (ii)

Adding (i) and (ii)

(a+b)⁴ + (a-b)⁷ = [⁴C₀a⁴ + ⁴C₁a³b + ⁴C₂a²b² + ⁴C₃a¹b³ + ⁴C₄b⁴] + [⁴C₀a⁴ - ⁴C₁a³b + ⁴C₂a²b² - ⁴C₃ab³ + ⁴C₄b⁴]

⇒ 2[⁴C₀a⁴ + ⁴C₂a²b² + ⁴C₄b⁴]

⇒ 2

⇒ 2[(1)a⁴ + (6)a²b² + (1)b⁴]

⇒ 2[a⁴ + 6a²b² + b⁴]

Therefore, (a+b)⁴ + (a-b)⁷ = 2[a⁴ + 6a²b² + b⁴]

Now, putting a = 2 and b = in the above equation.

= 2[(2)⁴ + 6(2)² ()² + ()⁴]

= 2(16+24x+x²)

Hence proved.