Solve -
3a = ( b + c + d )³
3b = (c + d + e )³
3c = (d + e + a )³
3d = (e + a + b)³
3e = (a + b + c)³
Given that the magnitude of the any three a , b, c , d, e doesn't lie in the interval (0, 1 )
Answers
Given : 3a = ( b + c + d )³
3b = (c + d + e )³
3c = (d + e + a )³
3d = (e + a + b)³
3e = (a + b + c)³
a , b, c , d, e doesn't lie in the interval (0, 1 )
To find : a , b , c , d & e
From the given equation symmetry its clear that
a = b = c = d = e
=> 3a = ( a + a + a)³
=> 3a = (3a)³
=> 3a = 27a³
=> a = 9a²
=> a (1 - 9a² ) = 0
=>a = 0 or 9a² = 1
=> a² = 1/9
=> a = ± 1/3
a = 0 , -1/3 , 1/3
a , b, c , d, e does not lies in [ 0 , 1/3]
hence a = - 1/3
a = b = c = d = e = -1/3
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Hola mate
Here is your answer -
From the given equation symmetry its clear that
a = b = c = d = e
=> 3a = ( a + a + a)³
=> 3a = (3a)³
=> 3a = 27a³
=> a = 9a²
=> a (1 - 9a² ) = 0
=>a = 0 or 9a² = 1
=> a² = 1/9
=> a = ± 1/3
a = 0 , -1/3 , 1/3
a , b, c , d, e does not lies in [ 0 , 1/3]
hence a = - 1/3