China | A Nice Algebra Problem | Math Olympiad

[@Quest3669]

a/b= 5/3 >0 from both eqns.
Putting this in 1st eqn
b^3= 27/ 8 or b= 3/ 2 and a= 5/2 soln.

[@RealQinnMalloryu4]

(12)+(13)=25 (2)+(3) (ab ➖ 3ab+2). (3)+(6) =9 (3)+(3^3) (1)+(3^1) (1)+(3) (ab ➖ 3ab+1).

[@wes9627]

a^2/25=b^2/9 or 3a=5b or b=(3/5)a;
a^3+(3/5)a^3=25; a^3=25*5/8=(5/2)^3, so a=5/2 and b=(3/5)(5/2)=3/2
Check: (5/2)^3+(5/2)^2*(3/2)=(125+75)/8=25 & (3/2)^3+(5/2)(3/2)^2=(27+45)/8=72/8=9

[@ganeshdas3174]

For a, b> 0 : a = 5 & b = 3

[@wes9627]

???

[@ganeshdas3174]

@wes9627 a/b ratio = 5/3

[@key_board_x]

a³ + a²b = 25
a².(a + b) = 25
a + b = 25/a² ← equation (1)
b³ + ab² = 9
b².(a + b) = 9
a + b = 9/b² → recall (1): a + b = 25/a²
9/b² = 25/a²
a²/b² = 25/9
(a/b)² = (± 5/3)²
a/b = ± 5/3 → given a > 0 and b > 0 → a/b > 0
a/b = 5/3
a = 5b/3 ← equation (2)
Restart
a³ + a²b = 25
b³ + ab² = 9
------------------------------------------------------------------------subtraction
(a³ + a²b) - (b³ + ab²) = 25 - 9
a³ + a²b - b³ - ab² = 16
a³ - b³ + a²b - ab² = 16
(a³ - b³) + (a²b - ab²) = 16
(a³ - b³) + ab.(a - b) = 16 → recall: a³ - b³ = (a - b).(a² + ab + b²)
(a - b).(a² + ab + b²) + ab.(a - b) = 16
(a - b).[(a² + ab + b²) + ab] = 16
(a - b).[a² + 2ab + b²] = 16
(a - b).(a + b)² = 16 → recall (2): a = 5b/3
[(5b/3) - b].[(5b/3) + b]² = 16
[(5b/3) - (3b/3)].[(5b/3) + (3b/3)]² = 16
(2b/3).(8b/3)² = 16
(2b/3).(64b²/9) = 16
128b³/27 = 16
b³ = (16 * 27)/128
b³ = 27/8
b³ = (9/2)³
→ b = 9/2
Recall: (2):
a = 5b/3
a = [5 * (9/2)]/3
a = [5 * 9/2]/3
a = 45/6
→ a = 15/2