# How can you factor #a^4 + b^4# and how can you apply it to the following examples ?

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#x^4+1#

#x^4+81#

#x^4+256#

##### 2 Answers

#### Explanation:

Notice that:

#(a^2-kab+b^2)(a^2+kab+b^2) = a^4+(2-k^2)a^2b^2+b^4#

So if we put

#a^4+b^4 = (a^2-sqrt(2)ab+b^2)(a^2+sqrt(2)ab+b^2)#

Hence we find:

#x^4+1 = (x^2-sqrt(2)x+1)(x^2+sqrt(2)x+1)" "# with#a=x# and#b=1#

#x^4+81 = (x^2-3sqrt(2)x+9)(x^2+3sqrt(2)x+9)" "# with#a=x# and#b=3#

#x^4+256 = (x^2-4sqrt(2)x+16)(x^2+4sqrt(2)x+16)" "# with#a=x# and#b=4#

**Historical note**

No less a mathematician than the great Leibniz thought that it was not possible to factor

This is no different, really, than the method already posted, but this is another way of thinking about the factorization process.

Note that

#a^4+b^4=(a^4+2a^2b^2+b^4)-2a^2b^2#

#=(a^2+b^2)^2-(sqrt2ab)^2#

Which is a factorable difference of squares:

#=((a^2+b^2)+sqrt2ab)((a^2+b^2)-sqrt2ab)#

#=(a^2+sqrt2ab+b^2)(a^2-sqrt2ab+b^2)#