Russian l can you solve?? l Hardest Olympiad Math Problem

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[@SuchetaDembowski]

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[@LEARN_WITH_SHAZAIN]

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[@dwiyantowibisana4928]

a=1, b=4

[@novellobrancher9627]

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[@joernbeineke]

a=4 and b=1 :)

[@lionelfischer8240]

ou a=1 et b =4, 2 solutions.

[@mtrichie111]

Lmao, first thing I thought before I played video

[@walterwen2975]

Hardest Olympiad Math Problem: a²(√a) + b²(√b) = 33, √a + √b = 3, a, b > 0; a, b =?
3 > √a > √b > 0 or 3 > √b > √a > 0, a, b ϵ R
√b = 3 - √a; a²(√a) + b²(√b) = (√a)⁵ + (√b)⁵ = (√a)⁵ + (3 - √a)⁵ = 33
Let: x = √a, (√a)⁵ + (3 - √a)⁵ = x⁵ + (3 - x)⁵ = x⁵ - (x - 3)⁵ = 33, a, b ϵ R; x ϵ R
x⁵ - [x⁵ - 5(3)x⁴ + 10(3²)x³ - 10(3³)x² + 5(3⁴)x - 3⁵] = 33
15(x⁴ - 6x³ + 18x² - 27x) + 243 - 33 = 15(x⁴ - 6x³ + 18x² - 27x + 14) = 0
x⁴ - 6x³ + 18x² - 27x + 14 = (x⁴ - 6x³ + 8x²) + (10x² - 27x + 14) = 0
(x - 2)(x³ - 4x²) + (x - 2)(10x - 7) = (x - 2)(x³ - 4x² + 10x - 7) = 0
x³ - 4x² + 10x - 7 = (x³ - 4x² + 3x) + 7(x - 1) = (x - 1)(x² - 3x) + 7(x - 1)
= (x - 1)(x² - 3x + 7), x⁴ - 6x³ + 18x² - 27x + 14 = (x - 2)(x - 1)(x² - 3x + 7) = 0
x² - 3x + 7 = 0, x = (3 ± i√19)/2, Rejected; x ϵ R
x - 2 = 0, x = 2 = √a, a = 4; √b = 3 - √a = 3 - 2 = 1, b = 1
x - 1 = 0, x = 1 = √a, a = 1; √b = 3 - √a = 3 - 1 = 2, b = 4
Answer check:;
a = 4, b = 1, a = 1, b = 4
a²(√a) + b²(√b) = 16(2) + 1 = 1 + 16(2) = 33, √a + √b = 2 + 1 = 1 + 2; Confirmed
Final answer:
a = 4, b = 1; a = 1, b = 4
Note:
Two complex value roots, if acceptable;
√a = (3 ± i√19)/2, a = [(3 ± i√19)/2]² = (- 5 ± 3i√19)/2; x = √a, x² - 3x + 7 = 0
√b = 3 - √a = (3 -/+ i√19)/2, b = (- 5 -/+ 3i√19)/2
Answer check:
a + 3√a + 7 = (- 5 ± 3i√19)/2 - 3[(3 ± i√19)/2] + 7 = - 14/2 + 7 = 0; Confirmed
√a + √b = (3 ± i√19)/2 + (3 -/+ i√19)/2 = 3; Confirmed

[@Марія-75]

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