How do you factor #27y^6-8#?

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Answer 1 · 2017-05-12T23:06:26

Rewrite this as

#27*y^6-8=(3y^2)^3-2^3#

Using the cubes difference formula

#a^3-b^3=(a-b)*(a^2+ab+b^2)#

where #a=3y^2# and #b=2#

we get

#27y^6-8=(3 y^2 - 2) (9 y^4 + 6 y^2 + 4)#

But this can be factored further as follows

#27y^6-8=1/3 (sqrt(6) - 3 y) (3 y + sqrt(6)) (-3 y^2 + sqrt(6) y - 2) (3 y^2 + sqrt(6) y + 2)#