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

## ${x}^{4} + 1$ ${x}^{4} + 81$ ${x}^{4} + 256$

Oct 10, 2016

${a}^{4} + {b}^{4} = \left({a}^{2} - \sqrt{2} a b + {b}^{2}\right) \left({a}^{2} + \sqrt{2} a b + {b}^{2}\right)$

#### Explanation:

Notice that:

$\left({a}^{2} - k a b + {b}^{2}\right) \left({a}^{2} + k a b + {b}^{2}\right) = {a}^{4} + \left(2 - {k}^{2}\right) {a}^{2} {b}^{2} + {b}^{4}$

So if we put $k = \sqrt{2}$ then we find:

${a}^{4} + {b}^{4} = \left({a}^{2} - \sqrt{2} a b + {b}^{2}\right) \left({a}^{2} + \sqrt{2} a b + {b}^{2}\right)$

Hence we find:

${x}^{4} + 1 = \left({x}^{2} - \sqrt{2} x + 1\right) \left({x}^{2} + \sqrt{2} x + 1\right) \text{ }$ with $a = x$ and $b = 1$

${x}^{4} + 81 = \left({x}^{2} - 3 \sqrt{2} x + 9\right) \left({x}^{2} + 3 \sqrt{2} x + 9\right) \text{ }$ with $a = x$ and $b = 3$

${x}^{4} + 256 = \left({x}^{2} - 4 \sqrt{2} x + 16\right) \left({x}^{2} + 4 \sqrt{2} x + 16\right) \text{ }$ with $a = x$ and $b = 4$

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Historical note

No less a mathematician than the great Leibniz thought that it was not possible to factor ${a}^{4} + {b}^{4}$ as a product of quadratics with real coefficients. So if you experienced any difficulty finding such a factorisation then you are in very good company.

Jun 11, 2017

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

Note that ${\left({a}^{2} + {b}^{2}\right)}^{2} = {a}^{4} + 2 {a}^{2} {b}^{2} + {b}^{4}$. We can then manipulate ${a}^{4} + {b}^{4}$ as such:

${a}^{4} + {b}^{4} = \left({a}^{4} + 2 {a}^{2} {b}^{2} + {b}^{4}\right) - 2 {a}^{2} {b}^{2}$

$= {\left({a}^{2} + {b}^{2}\right)}^{2} - {\left(\sqrt{2} a b\right)}^{2}$

Which is a factorable difference of squares:

$= \left(\left({a}^{2} + {b}^{2}\right) + \sqrt{2} a b\right) \left(\left({a}^{2} + {b}^{2}\right) - \sqrt{2} a b\right)$

$= \left({a}^{2} + \sqrt{2} a b + {b}^{2}\right) \left({a}^{2} - \sqrt{2} a b + {b}^{2}\right)$