2 3 2 y − 1 = y 3 + 1
Solve the equation above for real y . Assuming the roots are α , β , and γ , submit α + β + γ .
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Hi Mr. Alak, can you please elaborate on how you found the roots?
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3 2 y − 1 = 2 y 3 + 1
(Note that the Equation is of the form f ( f ( x ) ) = x ,or alternatively f ( x ) = f − 1 ( x ) , and since f ( x ) is an increasing function, f ( x ) = x , will provide solutions to the original equation, this can also be observed from the graph of f ( x ) = f − 1 ( x ) ).
x 3 − 2 x + 1 = 0 , x = 1 , 2 − 1 ± 5
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On removing the radical, the equation reduces to :
y 9 + 3 y 6 + 3 y 3 − 1 6 y + 9 = 0 .
The three real solutions of this equation are 1 , 2 5 − 1 , 2 − 5 − 1 , and their sum is 1 − 2 × 2 1 = 0 .