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

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

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

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

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