How do you write the expression #(y-3)(y-3)(y-3)# using exponents?

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Answer 1 · 2017-01-11T10:49:23

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

This is a product of three binomials, who are all the same i.e. #(y-3)(y-3)(y-3)#.

We can write it simply as #(y-3)^3#.

For actual mutiplication, we use distributive property, first over #(y-3)(y-3)# and then multiplying the result by #(y-3)#

#(y-3)(y-3)#

= #y(y-3)-3(y-3)#

= #yxxy-3xxy-3xxy-3xx(-3)#

= #y^2-3y-3y+9#

= #y^2-6y+9#

and now multiplying above again by #(y-3)#

#(y-3)(y-3)(y-3)#

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

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

= #yxxy^2-6yxxy+9xxy-3xxy^2-3xx(-6y)-3xx9#

= #y^3-6y^2+9y-3y^2+18y-27#

= #y^3-9y^2+27y-27#