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📅 May 3, 2026 👤 scraface

May 03, 2026

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Let \( y = f(x) \), so \( f(y) = x \). Then:
f(f(x)) = rac{(f(x))^3 - 3f(x)}{(f(x))^2 + 1} = x
Substitute \( f(x) = rac{x^3 - 3x}{x^2 + 1} \). Let \( N = x^3 - 3x \...
f(f(x)) = rac{(N/D)^3 - 3(N/D)}{(N/D)^2 + 1} = rac{ rac{N^3 - 3ND^2...
So, \( f(f(x)) = x \) becomes:
rac{N^3 - 3ND^2}{D(N^2 + D^2)} = x
Multiply both sides by \( D(N^2 + D^2) \):
N^3 - 3ND^2 = x D (N^2 + D^2)
Now substitute back \( N = x^3 - 3x \), \( D = x^2 + 1 \):
\( N^2 = (x^3 - 3x)^2 = x^6 - 6x^4 + 9x^2 \)

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