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Mathematics 21 Online
OpenStudy (anonymous):

Factor w^3+ 64 =

jimthompson5910 (jim_thompson5910):

w^3+64 (w)^3+(4)^3 (w+4)((w)^2-(w)(4)+(4)^2) (w+4)(w^2-4w+16) So w^3+64=(w+4)(w^2-4w+16)

jimthompson5910 (jim_thompson5910):

This is using the sum of cubes formula.

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