Question

Expand: `(sqrt(3) + sqrt(2))^4`

Answer 1

Here a = `sqrt(3)`, b = `sqrt(2)` and n = 4.

Using binomial theorem,

`(sqrt(3) + sqrt(2))^4`

`= ""^4"C"_0 (sqrt(3))^4 (sqrt(2))^0 + ""^4"C"_1 (sqrt(3))^3 (sqrt(2))^1 + ""^4"C"_2 (sqrt(3))^2 (sqrt(2))^2 + ""^4"C"_3 (sqrt(3))^1 (sqrt(2))^3 + ""^4"C"_4 (sqrt(3))^0 (sqrt(2))^4` 

Since ⁴C₀ = ⁴C₄ = 1, ⁴C₁ = ⁴C₃ = 4,

⁴C₂ = `(4 xx 3)/(2 xx 1)` = 6

 ∴ `(sqrt(3) + sqrt(2))^4`

`= 1(9) (1) + 4(3sqrt(3)) (sqrt(2)) + 6(3)(2) + 4(sqrt(3)) (2sqrt(2)) + 1(1)(4)`

= `9 + 12sqrt(6) + 36 + 8sqrt(6) + 4`

= `49 + 20sqrt(6)`