Question 232374
There are many factoring techniques. But this problem requires only one of them: the "Difference of squares" pattern, {{{a^2 - b^2 = (a+b)(a-b)}}}.<br>
We just have to realize what makes a term a perfect square. Numbers which are perfect squares are not hard to recognize. But variable factors which are perfect squares are not as obvious. For example, {{{x^8}}} would not seem to be a perfect square because 8 is not a perfect square. But {{{x^8}}} is a perfect square. In fact any variable with an even exponent is a perfect square! The reason for this is the rule for exponents: {{{a^(x*y) = (a^x)^y}}}. A specific case of this rule is: {{{a^(y*2) = (a^y)2}}}. On the left side we have an even exponent (since 2 times a whole number is even). On the right side we have something squared. (Similar logic tells us that {{{x^15}}} is a perfect cube because the exponent is divisible by 3, {{{x^20}}} is a perfect 4th power because the exponent is divisible by 4, etc.)<br>
Now we're ready to factor {{{x^8 - 1}}} as s difference of squares:
{{{x^8 - 1 = (x^4)^2 - 1^2 = (x^4+1)(x^4-1)}}}
When you reduce fractions, you always keep going until you can't go any further. Factoring is like reducing fractions because you keep going until you can't go any further. And if we look at {{{(x^4+1)(x^4-1)}}} and if we have learned what makes terms perfect squares, we'll realize that the second factor is another difference of squares:
{{{x^8 - 1 = (x^4)^2 - 1^2 = (x^4+1)(x^4-1) = (x^4+1)(x^2+1)(x^2-1)}}}
... and yet another difference of squares!!
{{{x^8 - 1 = (x^4)^2 - 1^2 = (x^4+1)(x^4-1) = (x^4+1)(x^2+1)(x^2-1)  = (x^4+1)(x^2+1)(x+1)(x-1)}}}
... and finally we're done. (Note: There is no "Sum of squares" pattern so those factors will not factor further.)