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

$\left({x}^{2} + {y}^{2}\right) \left({x}^{4} - {x}^{2} {y}^{2} + {y}^{4}\right) \left({x}^{12} - {x}^{6} {y}^{6} + {y}^{12}\right) + 2 {x}^{9} {y}^{9} = {\left({x}^{9} + {y}^{9}\right)}^{2}$

Oct 2, 2016

Using the sum of cubes identity

• ${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

the expansion of the square of a binomial

• ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

and the property of exponents

• ${\left({x}^{a}\right)}^{b} = {x}^{a b}$

we have

$\left({x}^{2} + {y}^{2}\right) \left({x}^{4} - {x}^{2} {y}^{2} + {y}^{4}\right) \left({x}^{12} - {x}^{6} {y}^{6} + {y}^{12}\right) + 2 {x}^{9} {y}^{9}$

$= \left[\left({x}^{2} + {y}^{2}\right) \left({\left({x}^{2}\right)}^{2} - {x}^{2} {y}^{2} + {\left({y}^{2}\right)}^{2}\right)\right] \left({x}^{12} - {x}^{6} {y}^{6} + {y}^{12}\right) + 2 {x}^{9} {y}^{9}$

$= \left({\left({x}^{2}\right)}^{3} + {\left({y}^{2}\right)}^{3}\right) \left({x}^{12} - {x}^{6} {y}^{6} + {y}^{12}\right) + 2 {x}^{9} {y}^{9}$

$= \left({x}^{6} + {y}^{6}\right) \left({\left({x}^{6}\right)}^{2} - {x}^{6} {y}^{6} + {\left({y}^{6}\right)}^{2}\right) + 2 {x}^{9} {y}^{9}$

$= {\left({x}^{6}\right)}^{3} + {\left({y}^{6}\right)}^{3} + 2 {x}^{9} {y}^{9}$

$= {x}^{18} + {y}^{18} + 2 {x}^{9} {y}^{9}$

$= {\left({x}^{9}\right)}^{2} + 2 {x}^{9} {y}^{9} + {\left({y}^{9}\right)}^{2}$

$= {\left({x}^{9} + {y}^{9}\right)}^{2}$