Question

1)$(a^{}frac{1}{3}-b^{frac{1}{3} } )*a^{}frac{2}{3} * b^{}frac{2}{3}$
2)$(frac{1}{x^{4} } -3)(frac{1}{x^{4} }+3)$
3)$(a^{1.5}-b^{0.5})^2$
4)$(p^{frac{1}{6}} + t^{frac{1}{8} })^2$

Answer 1

$1)(a^{frac{1}{3}}-b^{frac{1}{3}})*a^{frac{2}{3}}*b^{frac{2}{3}}=a^{frac{1}{3}+frac{2}{3}}*b^{frac{2}{3}}-a^{frac{2}{3}}*b^{frac{1}{3}+frac{2}{3}}=\\=a^{1}*b^{frac{2}{3}}-a^{frac{2}{3}}*b^{1}=ab^{frac{2}{3}}-a^{frac{2}{3}}b=ab^{frac{2}{3}}+(-a^{frac{2}{3}}b)$

$2)(frac{1}{x^4}-3)(frac{1}{x^4}+3)=(frac{1}{x^4})^2-3^2=frac{1}{x^8}-9$

$3)(a^{1,5}-b^{0,5})^2=(a^{1,5})^2-2*a^{1,5}*b^{0,5}+(b^{0,5})^2=\\=(a^{1,5*2})-2a^{1,5}b^{0,5}+(b^{0,5*2})=a^3-2a^1*a^{0,5}b^{0,5}+b^1=\\=a^3-2asqrt{ab}+b$

$4)(p^{frac{1}{6}}+t^{frac{1}{8}})^2=(p^{frac{1}{6}})^2+2*p^{frac{1}{6}}*t^{frac{1}{8}}+(t^{frac{1}{8}})^2=\\=p^{frac{2}{6}}+2p^{frac{1}{6}}t^{frac{1}{8}}+t^{frac{2}{8}}=p^{frac{1}{3}}+2p^{frac{1}{6}}t^{frac{1}{8}}+t^{frac{1}{4}}$