Разложите на множители:
1) $\frac{1}{64}a^8 - b^6$;
2) $a^3b^6c^9 + 8$;
3) $x^{21}y^{24} - m^{12}n^{15}$;
4) $a^6b^6 + 1$.
$\frac{1}{64}a^8 - b^6 = (\frac{1}{2})^6a^8 - b^6 = ((\frac{1}{2})^3a^4)^2 - (b^3)^2 = (\frac{1}{8}a^4 - b^3)(\frac{1}{8}a^4 + b^3)$
$a^3b^6c^9 + 8 = (ab^2c^3)^3 + 2^3 = (ab^2c^3 + 2)((ab^2c^3)^2 - 2ab^2c^3 + 2^2) = (ab^2c^3 + 2)(a^2b^4c^6 - 2ab^2c^3 + 4)$
$x^{21}y^{24} - m^{12}n^{15} = (x^{7}y^{8})^3 - (m^{4}n^{5})^3 = (x^{7}y^{8} - m^{4}n^{5})((x^{7}y^{8})^2 + x^{7}y^{8}m^{4}n^{5} + (m^{4}n^{5})^2) = (x^{7}y^{8} - m^{4}n^{5})(x^{14}y^{16} + x^{7}y^{8}m^{4}n^{5} + m^{8}n^{10})$
$a^6b^6 + 1 = (a^2b^2)^3 + 1^3 = (a^2b^2 + 1)((a^2b^2)^2 - a^2b^2 + 1) = (a^2b^2 + 1)(a^4b^4 - a^2b^2 + 1)$
Пожауйста, оцените решение
