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What is 64-z^6 factored down to?
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Is this correct? Help me verify this. ---> (2-z) (z^2+2z+4 2+z) (z^2-2z+4)
Ops I forgot to add a ()
\[2^{6} - z ^{6}\]
a^6 - b^6 = (a + b)(a^2 - ab + b^2)(a - b)(a^2 + ab + b^2)
plug that into there and done
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You got it?
Yeah I guess that's another way to solve it
That would be simplifying it fully
kk
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