Solution set of $\dfrac{(x + 1)^3 - (x - 1)^3}{(x + 1)^2 - (x - 1)^2} = 1$ comes from—

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Accepted Answer

simplifying to $\dfrac{6x^2 + 2}{4x} = 1$, then quadratic in $x$

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linear in $x$

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cubic in $x$

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an identity

Solution set of (x+1)3−(x−1)3(x+1)2−(x−1)2=1\dfrac{(x + 1)^3 - (x - 1)^3}{(x + 1)^2 - (x - 1)^2} = 1(x+1)2−(x−1)2(x+1)3−(x=1 comes from—