How do you prove the following identity using the factorization for sum of cubes without expanding either side?

#(x^2+y^2)(x^4-x^2y^2+y^4)(x^12-x^6y^6+y^12)+2x^9y^9=(x^9+y^9)^2#

1 Answer
Oct 2, 2016

Using the sum of cubes identity

  • #a^3+b^3 = (a+b)(a^2-ab+b^2)#

the expansion of the square of a binomial

  • #(a+b)^2 = a^2+2ab+b^2#

and the property of exponents

  • #(x^a)^b = x^(ab)#

we have

#(x^2+y^2)(x^4-x^2y^2+y^4)(x^12-x^6y^6+y^12)+2x^9y^9#

#=[(x^2+y^2)((x^2)^2-x^2y^2+(y^2)^2)] (x^12-x^6y^6+y^12)+2x^9y^9#

#=((x^2)^3+(y^2)^3)(x^12-x^6y^6+y^12)+2x^9y^9#

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

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

#=x^18+y^18+2x^9y^9#

#=(x^9)^2+2x^9y^9+(y^9)^2#

#=(x^9+y^9)^2#