a) $$ -x^3 + y^3 = y^3 - x^3 = (y - x)(y^2 + yx + x^2)$$.
б) $$ -a^6 + \frac{1}{8} = (\frac{1}{2})^3 - (a^2)^3 = (\frac{1}{2} - a^2)(\frac{1}{4} + \frac{1}{2}a^2 + a^4)$$.
в) $$ -8 - p^3 = -(8 + p^3) = -(2^3 + p^3) = -(2 + p)(4 - 2p + p^2)$$.
г) $$\frac{1}{-27} - b^6 = -(\frac{1}{27} + b^6) = - ((\frac{1}{3})^3 + (b^2)^3) = -(\frac{1}{3} + b^2)(\frac{1}{9} - \frac{1}{3}b^2 + b^4)$$.
д) $$x^6 + y^6 = (x^2)^3 + (y^2)^3 = (x^2 + y^2)(x^4 - x^2y^2 + y^4)$$.
Ответ: a) $$(y - x)(y^2 + yx + x^2)$$, б) $$(\frac{1}{2} - a^2)(\frac{1}{4} + \frac{1}{2}a^2 + a^4)$$, в) $$-(2 + p)(4 - 2p + p^2)$$, г) $$-(\frac{1}{3} + b^2)(\frac{1}{9} - \frac{1}{3}b^2 + b^4)$$, д) $$(x^2 + y^2)(x^4 - x^2y^2 + y^4)$$