Factors x^15+y^15
a)Factor x^15+y^15 by first considering it to be a sum of two cubes. You can three factors
b)Factor x^15+y^15 by first considering it to be a sum of fifth powers, Again, you van get three factors
c)Since the sets of factors you get in parts a and b are not identical, one more of the factors must not prime. By a clever application of techiniques you know, factor x^15+y^15 inti four polynomials with integer coefficients

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2014-09-19T20:34:55+05:30

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a)
 
 We\ know\ that \\ \\. \ \ \ \ \ a^3+b^3 = (a+b)(a^2-ab+b^2) ,\\ \\.\ \ \ \ \ x^5+y^5 = (x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)\\ \\ (x^5)^3 + (y^5)^3\\ \\ .\ \ \ = (x^5+y^5)(x^{10}-x^5y^5+y^{10}) \\.\ \ \ = (x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)(x^{10}-x^5y^5+y^{10}) \\ \\ \\ (x^3)^5+(y^3)^5 \\ \\ .\ \ \ \ = (x^3+y^3)(x^{12}-x^3y^9+x^6y^6-x^9y^3+y^{12}) \\ .\ \ \ \ = (x+y)(x^2-xy+y^2)(x^{12}-x^3y^9+x^6y^6-x^9y^3+y^{12}) \\ \\

So,

(x^4-x^3y+x^2y^2-xy^3+y^4)(x^{10}-x^5y^5+y^{10}) = \\. \ \ \ \ (x^2-xy+y^2)(x^{12}-x^3y^9+x^6y^6-x^9y^3+y^{12}) \\ \\

c)

(x^{10}-x^5y^5+y^{10})\\ = (x^2-xy+y^2)*(y^8+xy^7-x^3y^5-x^4y^4-x^5y^3+x^7y+x^8) \\ \\ 1-x^3y^9+x^6y^6-x^9y^3+y^{12} \\ = (y^8-y^7x+x^2y^6-x^3y^5+x^4y^4)* \\.\ \ \ \ \ \  (y^8+xy^7-x^3y^5-x^4y^4-x^5y^3+x^7y+x^8)\\ \\ \\ x^{15}+y^{15}\\ \\(x+y)(y^2-xy+x^2)(y^4-xy^3+x^2y^2-x^3y+x^4) * \\.\ \ \ \ \ \  (y^8+xy^7-x^3y^5-x^4y^4-x^5y^3+x^7y+x^8) \\ \\


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