How do you factor #y^3-3y^2+3y-1#?

2 Answers
Sep 23, 2015

Answer:

#y^3-3y^2+3y-1 = (y-1)^3#

Explanation:

The 1, 3, 3, 1 pattern is a giveaway.

In general, #(a+b)^3 = a^3+3a^2b+3ab^2+b^3#

More generally, you can expand #(a+b)^n# or recognise an expanded power of #(a+b)^n# by looking at the appropriate row of Pascal's triangle:

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For example:

#(a+b)^5 = a^5+5a^4b^3+10a^3b^2+10a^2b^3+5ab^4+b^5#

Sep 23, 2015

Answer:

#y^3-3y^2+3y-1 = (y-1)^3#

Explanation:

A quick inspection reveals that if #y=1# then
#color(white)("XXX")y^3-3y^2+3y-1 = 0#

Therefore #(y-1)# is a factor of #y^3-3y^2+3y-1#

#(y^3-3y^2+3y-1)div(y-1) = y^2-2y+1#
and
#y^2-2y+1 = (y-1)(y-1)#

Therefore
#y^3-3y^2+3y-1#
#color(white)("XXX")=(y-1)(y-1)(y-1)#
#color(white)("XXX")=(y-1)^3#

Note that this could also have been solved if you recognized the pattern of the given expression as a row from Pascal's triangle, but I thought a more general method of solution might be useful.