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How do you factor #y^3-64#?

1 Answer

Jan 30, 2016

#y^3-4^3=(y-4)(y^2+4y+16)#

Explanation:

#y^3-64# is a difference of cubes, #a^3– b^3 = (a – b)(a^2 + ab + b^2)#, where #a=y# and #b=4#.

Rewrite the equation.

#(y)^3-(4)^3=(y-4)(y^2+(y*4)+4^2)#

Simplify.

#y^3-4^3=(y-4)(y^2+4y+16)#

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