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stupidtrash
20.08.2019 •
Mathematics
Is the hcf of a^3-8, a^4+16 and a^3+2a^2+4a just 1?
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Ответ:
a^3-8 is the differnce between 2 perfect cubes and is factored out to
(a-2)(a^2+2a+4)
the factors are (a-2) and (a^2+2a+4)
a^4+16
this is not factorable, unless you mistiped it and it was really a^4-16
the factor is 1
a^3+2a^2+4a
factor out the a
a(a^2+2a+4)
this is the most factored out form
the factors are (a) and (a^2+2a+4)
there are no common factors except 1 so 1 is the answer
Ответ:
No, but maybe we can play among us later?
Step-by-step explanation:
Im not sus i swear lol