A Nice Math Olympiad Problem (x+2)(x+3)(x+4)=990 | Best Trick... | Math Olympiad

[@michaeldoerr5810]

That is one of the best vids that I have seen. And yhis might be how I tackle other quartic equations!!!

[@key_board_x]

(x + 2).(x + 3).(x + 4) = 990
(x + 2).(x + 2 + 1).(x + 2 + 2) = 990 → let: y = x + 2
y.(y + 1).(y + 2) = 990
y.(y² + 2y + y + 2) = 990
y.(y² + 3y + 2) = 990
y³ + 3y² + 2y - 990 = 0 → let: y = p - 1 ← to make item to the 2nd power disappeared
(p - 1)³ + 3.(p - 1)² + 2.(p - 1) - 990 = 0
(p - 1)².(p - 1) + 3.(p² - 2p + 1) + 2p - 2 - 990 = 0
(p² - 2p + 1).(p - 1) + 3p² - 6p + 3 + 2p - 2 - 990 = 0
p³ - p² - 2p² + 2p + p - 1 + 3p² - 6p + 3 + 2p - 2 - 990 = 0
p³ - p - 990 = 0
p³ - p - (1000 - 10) = 0
p³ - p - 1000 + 10 = 0
p³ - 1000 - p + 10 = 0
(p³ - 10³) - (p - 10) = 0 → recall: a³ - b³ = (a - b).(a² + ab + b²)
(p - 10).(p² + 10p + 100) - (p - 10) = 0
(p - 10).[(p² + 10p + 100) - 1] = 0
(p - 10).(p² + 10p + 100 - 1) = 0
(p - 10).(p² + 10p + 99) = 0
First case: (p - 10) = 0
p - 10 = 0
p = 10 → recall: y = p - 1
y = 9 → recall: y = x + 2 → x = y - 2
x = 7
Second case: (p² + 10p + 99) = 0
p² + 10p + 99 = 0
Δ = 10² - (4 - 99) = 100 - 396 = - 296 = 296i² = 4i² * 74 = (2i)² * 74
p = (- 10 ± 2i√74)/2
p = - 5 ± i√74 → recall: y = p - 1
y = - 6 ± i√74 → recall: x = y - 2
x = - 8 ± i√74

[@olocomeu7137]

999 or 990???? 😂😂😂😂