A cubic polynomial $$ ax^3 + bx^2 + cx + d $$ intersects the x-axis at $$ x = 1 $$ and $$ x = -1 $$, and the y-axis at $$ y = 2 $$. Determine the value of $$ b $$.

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The given polynomial is $$ ax^3 + bx^2 + cx + d $$. Since the polynomial has roots at $$ x = 1 $$ and $$ x = -1 $$, it must include the factor $$ (x - 1)(x + 1) = x^2 - 1 $$. Thus, the polynomial can be expressed as:

$$
ax^3 + bx^2 + cx + d = a(x^3 - x) + bx^2 + cx + d
$$

It is also given that the polynomial intersects the y-axis at $$ y = 2 $$, which means when $$ x = 0 $$, $$ d = 2 $$.

Now, substituting $$ x = 1 $$ and $$ x = -1 $$ into the polynomial, we set up equations for these roots:

At $$ x = 1: a(1^3 - 1) + b(1^2) + c(1) + d = 0 $$

At $$ x = -1: a((-1)^3 - (-1)) + b((-1)^2) + c(-1) + d = 0 $$

Substituting $$ d = 2 $$ into these equations and solving for the coefficients using algebraic methods, we find:

$$
b = -2, c = 0, d = 2
$$

Thus, the value of $$ b $$ is **$$ -2 $$**.

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