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how do you factor 27y^3+8?
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\(\huge a^3+b^3=(a+b)(a^2-ab+b^2)\)
and here let a=3y and b=2
write 27y^3 as(3y)^3 and ^^
alright but why is a=3 and b=2?
nopes, a=3y, not 3
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and u had 8, u can write that as 2^3
so 27y^3+8= (3y)^3+2^3 got this?
yea so u just plug it into the equation and get (3+2)(3^2-(3)(2)+2^2) right?....
NOT 3+2 3y+2 a=3y
alright so then u just solve- so i got (3y+2)(9y^2-6y+4) i this correct?
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yes! that is correct :)
kool thanks!
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