Упр.399 ГДЗ Никольский Потапов 9 класс (Алгебра)
а) (x^(1/3)-y^(1/2))^3
б) (m^(1/2)+n^(2/3))^3
в) (a^(1/2)-b^(1/3))^3
Используем формулу куба суммы и куба разности:
$$ (u-v)^3=u^3-3u^2v+3uv^2-v^3, $$
$$ (u+v)^3=u^3+3u^2v+3uv^2+v^3. $$
$$ \left(x^{\frac13}-y^{\frac12}\right)^3 =\left(x^{\frac13}\right)^3-3\left(x^{\frac13}\right)^2y^{\frac12}+3x^{\frac13}\left(y^{\frac12}\right)^2-\left(y^{\frac12}\right)^3 $$
$$ =x-3x^{\frac23}y^{\frac12}+3x^{\frac13}y-y^{\frac32}. $$
$$ \left(m^{\frac12}+n^{\frac23}\right)^3 =\left(m^{\frac12}\right)^3+3\left(m^{\frac12}\right)^2n^{\frac23}+3m^{\frac12}\left(n^{\frac23}\right)^2+\left(n^{\frac23}\right)^3 $$
$$ =m^{\frac32}+3mn^{\frac23}+3m^{\frac12}n^{\frac43}+n^2. $$
$$ \left(a^{\frac12}-b^{\frac13}\right)^3 =\left(a^{\frac12}\right)^3-3\left(a^{\frac12}\right)^2b^{\frac13}+3a^{\frac12}\left(b^{\frac13}\right)^2-\left(b^{\frac13}\right)^3 $$
$$ =a^{\frac32}-3ab^{\frac13}+3a^{\frac12}b^{\frac23}-b. $$
Ответ
а) $$x-3x^{\frac23}y^{\frac12}+3x^{\frac13}y-y^{\frac32}$$; б) $$m^{\frac32}+3mn^{\frac23}+3m^{\frac12}n^{\frac43}+n^2$$; в) $$a^{\frac32}-3ab^{\frac13}+3a^{\frac12}b^{\frac23}-b$$.
