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Factor the polynomial completely 27-y^3
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There is a pretty simple formula for the difference of square which is \[a^3 + b^3 = (a - b)(a^2 +ab + b^2)\] so a = y and b = 27. Try it from there
would y =1?
no y is just a general variable factoring isn't trying to find the values of y. Since this is just an EXPRESSION not an EQUATION y could be any number because there are no restrictions
so it would be (y-27)(y^2+27y+27^2)?? then would i FOIL?
nope this is the factorization of the original expression
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so what would i do after this step?
27-y^(3) The binomial can be factored using the difference of cubes formula, because both terms are perfect cubes. The difference of cubes formula is a^(3)-b^(3)=(a-b)(a^(2)+ab+b^(2)). (3-y)(9+3y+y^(2))
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