<s>[INST] How many roots does $(1+z^3)^8 = (1 + z^4)^6$ have? I got stuck on this question: 
How many roots does $(1+z^3)^8 = (1 + z^4)^6$ have? (including complex roots and roots with multiplicity)
My attempt at a solution:
First we can write the equation as:
$$(1+z^3)^8 - (1 + z^4)^6 = 0$$
We can then factor $(1+z^3)$
$$(z+1)^8(z^2-z+1)^8 - (1 + z^4)^6 = 0$$
But I am stuck here, the expression $(1+z^4)$ cannot easily be factored, but it should have 4 complex roots. I think there should be a simpler way to solve this question...
 [/INST] $(z^3+1)^8 = (z^4+1)^6$
$z^{24} + 8z^{21} + 28z^{18} + 56z^{15} + 70z^{12} + 56z^9 + 28z^6 + 8z^3 + 1 = z^{24} + 6z^{20} + 15z^{16} + 20z^{12} + 15z^8 + 6z^4 + 1$
$z^{24} + 8z^{21} + 28z^{18} + 56z^{15} + 70z^{12} + 56z^9 + 28z^6 + 8z^3 + 1 - z^{24} - 6z^{20} - 15z^{16} - 20z^{12} - 15z^8 - 6z^4 - 1 = 0$
$z^3(8z^{18} - 6z^{17} + 28z^{15} - 15z^{13} + 56z^{12} + 50z^9 + 56z^6 - 15z^5 + 28z^3 - 6z + 8) = 0$
$z^3(4z^6 - 3z^5 + 6z^3 - 3z + 4)(2z^{12} + 4z^9 + 3z^8 + 6z^6 + 3z^4 + 4z^3 + 2)$
Cannot be factored further, but we can try to solve it from there:

*

*$z^3 = 0 \Rightarrow \sqrt[3]{z^3} = \sqrt[3]{0} \Rightarrow z = 0$

*$4z^6 - 3z^5 + 6z^3 - 3z + 4 = 0$ has 6 complex roots (three pairs of complex conjugates)

*$2z^{12} + 4z^9 + 3z^8 + 6z^6 + 3z^4 + 4z^3 + 2 = 0$ has 12 complex roots (six pairs of complex conjugates)

The equation has 19 roots.
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