a) $$100a^4 - \frac{1}{9}b^2 = (10a^2)^2 - (\frac{1}{3}b)^2 = (10a^2 - \frac{1}{3}b)(10a^2 + \frac{1}{3}b)$$
б) $$9x^2 - (x - 1)^2 = (3x)^2 - (x - 1)^2 = (3x - (x - 1))(3x + (x - 1)) = (3x - x + 1)(3x + x - 1) = (2x + 1)(4x - 1)$$
в) $$x^3 + y^6 = x^3 + (y^2)^3 = (x + y^2)(x^2 - xy^2 + y^4)$$
Ответ:
a) $$(10a^2 - \frac{1}{3}b)(10a^2 + \frac{1}{3}b)$$
б) $$(2x + 1)(4x - 1)$$
в) $$(x + y^2)(x^2 - xy^2 + y^4)$$