Math Olympiad | A Nice Exponential Problem | 90% Failed to solve! 

((Sqrt[39]+Sqrt[3])/Sqrt[12])^7=(337+97Sqrt[13])/2=168.5+48.5Sqrt[13] It’s in my head.

(Sqrt(39)+sqrt(3))/sqrt(12) = (sqrt(13)+1)/2, so the we are surching ((sqrt(13)+1)^7)/128. That's simpler. Now, (sqrt(13)+1)^2 = 14+2.sqrt(13); (sqrt(13)+1)^4 = 196+52+56.sqrt(13)= 248+56.sqrt(13)=8.(31+7.sqrt(13)) (a) sqrt(13+1)^3 = 26+14+16.sqrt(13) = 40+16.sqrt(13) = 8.(5+2.sqrt(13) (b). Then we multiply (a) and (b): (sqrt(13)+1)^7 = 64.(155+182+97.sqrt(13)) = 64.(337+97.sqrt(13), then finally the result is (337+97.sqrt(13))/2.

Why do you calculate the approximate value of x^7 and check ? we only simplify (1101+291√13)/6 = 337 +97√13 and stop. that's okay.

Let x = (√39 + √3) / √12 after simplification x = (1 + √13)/2 x^2 = ((1 + √13)/2)^2 = (14 + 2√13)/4 = (7 + √13)/2 = x + 3 x^3 = x^2 + 3*x = 4*x + 3 x^4 = 4*x^2 + 3*x = 7*x + 12 x^7 = x^4 * x^3 = (7*x + 12)* (4*x + 3) = 28*x^2 + 21*x + 48*x + 36 x^7 = 28*x^2 + 69*x + 36 = 28*(x+3) + 69*x + 36 = 97*x + 120 x^7 = 97* (1 + √13)/2 + 120 = (337 + 97√13) / 2

Cannot apprciate the method. Direct expansion would have taken less time without straining the mind😢

小数に直すなんてあり得ない。それだったら最初から小数に直して７乗すればいい。

[(√39 + √3)/√12]² = (√39 + √3)²/(√12)² [(√39 + √3)/√12]² = [39 + 2√(39 * 3) + 3]/12 [(√39 + √3)/√12]² = [42 + 2√(3 * 13 * 3)]/12 [(√39 + √3)/√12]² = [42 + 6√13]/12 [(√39 + √3)/√12]² = (7 + √13)/2 [(√39 + √3)/√12]⁴ = { [(√39 + √3)/√12]² }² [(√39 + √3)/√12]⁴ = { (7 + √13)/2 }² [(√39 + √3)/√12]⁴ = (7 + √13)²/4 [(√39 + √3)/√12]⁴ = (49 + 14√13 + 13)/4 [(√39 + √3)/√12]⁴ = (62 + 14√13)/4 [(√39 + √3)/√12]⁴ = (31 + 7√13)/2 [(√39 + √3)/√12]⁷ = [(√39 + √3)/√12]⁴.[(√39 + √3)/√12]².[(√39 + √3)/√12] [(√39 + √3)/√12]⁷ = [(31 + 7√13)/2] * [(7 + √13)/2].[(√39 + √3)/√12] [(√39 + √3)/√12]⁷ = (31 + 7√13).(7 + √13).(√39 + √3)/(2 * 2 * √12) [(√39 + √3)/√12]⁷ = (217 + 31√13 + 49√13 + 91).(√39 + √3)/(2 * 2 * 2√3) [(√39 + √3)/√12]⁷ = (308 + 80√13).(√39 + √3)/(2 * 2 * 2√3) [(√39 + √3)/√12]⁷ = (77 + 20√13).(√39 + √3)/(2√3) [(√39 + √3)/√12]⁷ = [77√39 + 77√3 + 20√(13 * 39) + 20√(13 * 3)]/(2√3) [(√39 + √3)/√12]⁷ = [77√(3 * 13) + 77√3 + 20√(13 * 3 * 13) + 20√(13 * 3)]/(2√3) → simpification by √3 [(√39 + √3)/√12]⁷ = [77√13 + 77 + 20√(13 * 13) + 20√13]/2 [(√39 + √3)/√12]⁷ = [97√13 + 77 + 260]/2 [(√39 + √3)/√12]⁷ = [97√13 + 337]/2