How do find the quotient of #(y^3 - 125)/(y - 5)#?

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Answer 1 · 2015-10-30T05:27:48

#(y^2+5y+25)#

You need to factor in order to get rid of the denominator. In order to do that you need to use the difference of cubes. That is

#a3 + b3 = (a + b)(a2 – ab + b2) #

#a3 – b3 = (a – b)(a2 + ab + b2)#

We are going to use the second one

Also #a=y, b=5#

Turn #y^3 -125# to #y^3 - 5^3#

Use the formula

#y^3 - 5^3=(y-5)(y^2+5y+25)#

Plug that into the numerator

#((y-5)(y^2+5y+25))/((y-5))#

Simplify

#(y^2+5y+25)#