Question

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`

Answer 1

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`