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Math Olympiad | A Nice Exponential Problem | VIJAY Maths

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VIJAY Maths

VIJAY Maths

Ай бұрын

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Пікірлер: 11
@SOLVINGMATHPROBLEMS
@SOLVINGMATHPROBLEMS Ай бұрын
How you could decide that 1296 is to be written as 729+567?
@denizaydemir6756
@denizaydemir6756 Ай бұрын
I think experience makes it happen, he already knew which formulas are going to give the answer so he manipulated the equation in a way that’d be working for him, that is my guess
@andryvokubadra2644
@andryvokubadra2644 Ай бұрын
1296 = 6*6³ 1000 = 10³ 1331 = 11³ he just guessed whatever possible of 1296 729 + 567 = 1296 9³ + 9²*7 = 1296 9²(9+7) = 1296 9²*16 = 1296 9²*4² = 1296 Some possible root : 1296 = 6²*6² 1296 = 6*6³ (unperfect cubic) 1296 = 4²*9² (he is) 1296 = 4*9²*4 And just 9 correct 😀😀😀 Becouse 1296 is not a perfect cubic, all never easy becouse just 1 or 2 real root. other is imaginary one 😅😅😅
@andryvokubadra2644
@andryvokubadra2644 Ай бұрын
1296 = 6*6³ 1000 = 10³ 1331 = 11³ he just guessed whatever possible of 1296 729 + 567 = 1296 9³ + 9²*7 = 1296 9²(9+7) = 1296 9²*16 = 1296 9²*4² = 1296 Some possible root : 1296 = 6²*6² 1296 = 6*6³ (unperfect cubic) 1296 = 4²*9² (he is) 1296 = 4*9²*4 Pattern of (9³ + 9²*7) similiarry (x³ + 7x²) 😀😀😀 Becouse 1296 is not a perfect cubic, all never easy becouse just 1 or 2 real root. other is imaginary one 😅😅😅
@gaiatetuya92
@gaiatetuya92 Ай бұрын
1296=329+567 は 何処から出て来た?
@vijaymaths5483
@vijaymaths5483 Ай бұрын
Not 329 i. e 729
@ezzatabdo5027
@ezzatabdo5027 Ай бұрын
Just oversight, welcome
@skibidi.G
@skibidi.G Ай бұрын
Thank you , very nice, but no highschool students will ever split 1296 like that 😔
@andryvokubadra2644
@andryvokubadra2644 Ай бұрын
@@skibidi.G Becouse the basic pattern is : ax³ + bx² + cx + d = 0 Then we can expand : (x+x1)(ax² + bx + c) = 0 (x+x1)(x+x2)(x+x3) = 0 But. . . when it've lost a part (bx² example) and C or D not the perfect cubic number. . . tralala. . . a big problem 😁😁😁🤣🤣🤣
@skibidi.G
@skibidi.G Ай бұрын
@@andryvokubadra2644 oh interesting 😊
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