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Mathematics 8 Online
OpenStudy (anonymous):

how do you factor 27y^3+8?

hartnn (hartnn):

\(\huge a^3+b^3=(a+b)(a^2-ab+b^2)\)

OpenStudy (anonymous):

and here let a=3y and b=2

hartnn (hartnn):

write 27y^3 as(3y)^3 and ^^

OpenStudy (anonymous):

alright but why is a=3 and b=2?

hartnn (hartnn):

nopes, a=3y, not 3

hartnn (hartnn):

and u had 8, u can write that as 2^3

hartnn (hartnn):

so 27y^3+8= (3y)^3+2^3 got this?

OpenStudy (anonymous):

yea so u just plug it into the equation and get (3+2)(3^2-(3)(2)+2^2) right?....

hartnn (hartnn):

NOT 3+2 3y+2 a=3y

OpenStudy (anonymous):

alright so then u just solve- so i got (3y+2)(9y^2-6y+4) i this correct?

hartnn (hartnn):

yes! that is correct :)

OpenStudy (anonymous):

kool thanks!

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