Упростите (1/3x^2 + 1/6y^2)(1/9x^4 - 1/18x^2y^2 + 1/36y^4) - 1/216y^6

Let the expression be E.

E = (\(\frac{1}{3}x^2 + \frac{1}{6}y^2\))(\(\frac{1}{9}x^4 - \frac{1}{18}x^2y^2 + \frac{1}{36}y^4\)) - \(\frac{1}{216}y^6\)

Recognize the second factor as a sum of cubes pattern: \(a^2 - ab + b^2\) where \(a = \frac{1}{3}x^2\) and \(b = \frac{1}{6}y^2\).

So, \((\frac{1}{3}x^2)^3 + (\frac{1}{6}y^2)^3 = \frac{1}{27}x^6 + \frac{1}{216}y^6\).

Therefore, E = \(\frac{1}{27}x^6 + \frac{1}{216}y^6\) - \(\frac{1}{216}y^6\) = \(\frac{1}{27}x^6\).