Solving the Polynomial Equation

We are given the equation: $(x + 1)(x + p)(x^2 + p^2) = x^4 - 1$ Our goal is to find the value of the variable $p$.

Factoring the Right Side

The term $x^4 - 1$ on the right side of the equation can be factored using the difference of squares formula, $a^2 - b^2 = (a - b)(a + b)$. Here, $a = x^2$ and $b = 1$. $x^4 - 1 = (x^2)^2 - 1^2 = (x^2 - 1)(x^2 + 1)$ Furthermore, the term $x^2 - 1$ is also a difference of squares, where $a = x$ and $b = 1$. So, $x^2 - 1 = (x - 1)(x + 1)$. Substituting this back, we get: $x^4 - 1 = (x - 1)(x + 1)(x^2 + 1)$

Comparing Both Sides

Now, we can rewrite the original equation with the factored right side: $(x + 1)(x + p)(x^2 + p^2) = (x - 1)(x + 1)(x^2 + 1)$ For this equation to hold true for all values of $x$, the expressions on both sides must be equivalent. If we consider values of $x$ where $x \neq -1$, we can cancel the $(x + 1)$ term from both sides: $(x + p)(x^2 + p^2) = (x - 1)(x^2 + 1)$ By comparing the structure of the factors on both sides, we can see a direct match if we set $p = -1$. Let's substitute $p = -1$ into the left side expression: $(x + (-1))(x^2 + (-1)^2) = (x - 1)(x^2 + 1)$ This resulting expression exactly matches the right side $(x - 1)(x^2 + 1)$.

Verification

Let's substitute $p = -1$ back into the original equation to verify: $(x + 1)(x + (-1))(x^2 + (-1)^2) \stackrel{?}{=} x^4 - 1$ $(x + 1)(x - 1)(x^2 + 1) \stackrel{?}{=} x^4 - 1$ Using the difference of squares formula $(a+b)(a-b)=a^2-b^2$: $(x^2 - 1)(x^2 + 1) \stackrel{?}{=} x^4 - 1$ Again, using the difference of squares formula: $(x^2)^2 - 1^2 \stackrel{?}{=} x^4 - 1$ $x^4 - 1 = x^4 - 1$ The equation holds true. Therefore, the value of $p$ must be $-1$.