Question

Розклади на множники:

256−z32.


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Answer 1

$256-z^{32}=16^{2}-(z^{16})^{2}=(16-z^{16})(16+z^{16})=[4^{2}-(z^{8})^{2}](16+z^{16})=\\\\=(4-z^{8})(4+z^{8})(16+z^{16})=[2^{2}-(z^{4} )^{2}](4+z^{8})(16+z^{16})=\\\\=\boxed{(2-z^{4})(2+z^{4})(4+z^{8})(16+z^{16})}$