Ответ: смотри решение ниже
\[x^3 - y^3 = (x - y)(x^2 + xy + y^2)\]
\[a^3 - 8 = (a - 2)(a^2 + 2a + 4)\]
\[b^3 - 27c^3 = (b - 3c)(b^2 + 3bc + 9c^2)\]
\[8d^3 - 64k^3 = 8(d^3 - 8k^3) = 8(d - 2k)(d^2 + 2dk + 4k^2)\]
\[125m^3 - 27n^6 = (5m - 3n^2)(25m^2 + 15m \cdot n^2 + 9n^4)\]
\[b^3 + 0.001e^3 = (b + 0.1e)(b^2 - 0.1be + 0.01e^2)\]
\[p^3 + \frac{8}{27}k^3 = (p + \frac{2}{3}k)(p^2 - \frac{2}{3}pk + \frac{4}{9}k^2)\]
\[\frac{64}{125}m^3 + 8 = \frac{8}{125}(8m^3 + 125) = \frac{8}{125}(2m + 5)(4m^2 - 10m + 25)\]
\[t^9 + f^{12} = (t^3 + f^4)(t^6 - t^3f^4 + f^8)\]
\[25x^2y^2 - 16 = (5xy - 4)(5xy + 4)\]
\[a^6 - 4a^3b + 4b^2 = (a^3 - 2b)^2\]
\[49a^2 - 112ab + 64b^2 = (7a - 8b)^2\]
\[0.81x^{10} - 400y^{12} = (0.9x^5 - 20y^6)(0.9x^5 + 20y^6)\]
\[25m^{10} + n^8 + 10m^5n^4 = (5m^5 + n^4)^2\]
\[-1 + 64m^4b^6 = (8m^2b^3 - 1)(8m^2b^3 + 1)\]
\[\frac{1}{64}n^6 - 27m^3 = (\frac{1}{4}n^2 - 3m)(\frac{1}{16}n^4 + \frac{3}{4}mn^2 + 9m^2)\]
\[\frac{1}{125}x^9 + 0.08y^6 = (\frac{1}{5}x^3 + 0.2y^2)(\frac{1}{25}x^6 - \frac{1}{25}x^3y^2 + 0.04y^4)\]
\[\frac{1}{9}m^2n^2 - \frac{1}{25}a^{11}b^2 = (\frac{1}{3}mn - \frac{1}{5}a^{\frac{11}{2}}b)(\frac{1}{3}mn + \frac{1}{5}a^{\frac{11}{2}}b)\]
\[0.01x^2 + 2xy + 100y^2 = (0.1x + 10y)^2\]
Ответ: смотри решение ниже
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