dddan
how do you factor 27y^3+8?



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

mukushla
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and here let a=3y and b=2

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

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

hartnn
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nopes, a=3y, not 3

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

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

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

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

dddan
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alright so then u just solve so i got (3y+2)(9y^26y+4) i this correct?

hartnn
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yes!
that is correct :)

dddan
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kool thanks!