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I4talent Method 1: $\displaystyle f(x) = x^4-1$
$\displaystyle f(1) = (1)^4-1 = 1-1 = 0$, therefore $\displaystyle (x-1)$ is a factor of $\displaystyle x^4-1$.
$\displaystyle f(-1) = (-1)^4-1= 1-1 = 0$, therefore $\displaystyle (x+1)$ is a factor of $\displaystyle x^4-1$.
So we have $\displaystyle x^4-1 = (x-1)(x+1)(ax^2+bx+c) $
$\displaystyle (x^2-1)(ax^2+bx+c) = x^4-1$
$\displaystyle x^2(ax^2+bx+c)-1(ax^2+bx+c) = x^4-1$
$\displaystyle ax^4+bx^3+cx^2-ax^3-bx^2-cx = x^4-1$
$\displaystyle (a)x^4+(b)x^3+(c-a)x^2-(b)x-c = x^4-1$
$\displaystyle \begin{cases} a = 1 \\
b = 0 \\
c-a = 0 & c-1 = 0 \implies c = 1 \\
-b = 0 \\
-c = -1 & c = 1. \end{cases}$
Therefore $\displaystyle ax^2+bx+c = x^2+1$
Hence $\displaystyle \boxed{(x^2-1)(x^2+1) = x^4-1}$
Method 2: Any products yielding answers of the form $\displaystyle a^n-b^n$ are given by:
$\displaystyle (a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+...+ab^{n-1}+b^{n-1}) = a^n-b^n$
So for $\displaystyle x^4-1^2$, we have:
$\displaystyle (x-1)((x^{4-1})+(x^{4-2})({1})+(x^{4-3})({1})^2+(x^{4-4})({1})^3)= x^4-1^2$
$\displaystyle (x-1)((x^{3})+(x^{2})({1})+(x^{1})({1})^2+(x^{0})({1} )^3)= x^4-1^2$
$\displaystyle (x-1)(x^3+x^2+x+1) = x^4-1$
$\displaystyle x^3+x^2+x+1$ yields to zero when $\displaystyle x = -1$, so, by the factor theorem:
$\displaystyle (x+1)(x-1)(ax^2+bx+c) = x^4-1$
$\displaystyle (x^2-1)(ax^2+bx+c) = x^4-1$
$\displaystyle x^2(ax^2+bx+c)-1(ax^2+bx+c) = x^4-1$
$\displaystyle ax^4+bx^3+cx^2-ax^3-bx^2-cx = x^4-1$
$\displaystyle (a)x^4+(b)x^3+(c-a)x^2-(b)x-c = x^4-1$
$\displaystyle \begin{cases} a = 1 \\
b = 0 \\
c-a = 0 & c-1 = 0 \implies c = 1 \\
-b = 0 \\
-c = -1 & c = 1. \end{cases}$
Therefore $\displaystyle ax^2+bx+c = x^2+1$
Hence $\displaystyle \boxed{(x^2-1)(x^2+1) = x^4-1}$