Dla n=3,4,6,9,12 oblicz n-te potęgi podanego pierwiastka. a) (kwadratowy pierwiastek z 2)do potęgi n b) (kwadratowy pierwiastek z 2/3)do potęgi n c) (pierwiastek sześcienny z 7)do potęgi n d) (pierwiastek sześcienny z 3/2) do potęgi n BARDZO PROSZĘ

[latex]a)\ (sqrt{2})^3 = 2sqrt{2}\ (sqrt{2})^4=[(sqrt{2})^2]^2=2^2= 4\ (sqrt{2})^6 =[(sqrt{2})^2]^3 = 2^3=8\ (sqrt{2})^9 = (sqrt{2})^8 cdot sqrt{2}=[ (sqrt{2})^2]^4cdot sqrt{2}=2^4cdot sqrt{2}=16sqrt{2}\ (sqrt{2})^{12} = [(sqrt{2})^6]^2 = 8^2=64[/latex] [latex]b)\ (sqrt{frac{2}{3}})^3 = frac{2sqrt{2}}{3sqrt{3}}=frac{2sqrt{6}}{9}\ (sqrt{frac{2}{3}})^4 = [(sqrt{frac{2}{3}})^2]^2 = (frac{2}{3})^2 = frac{4}{9}\ (sqrt{frac{2}{3}})^6 = [(sqrt{frac{2}{3}})^2]^3 = (frac{2}{3})^3 = frac{8}{27}\ (sqrt{frac{2}{3}})^9 =(sqrt{frac{2}{3}})^8cdot sqrt{frac{2}{3}}= [(sqrt{frac{2}{3}})^2]^4=(frac{2}{3})^4cdot sqrt{frac{2}{3}}= frac{16sqrt{2}}{81sqrt{3}}=frac{16sqrt{6}}{243}\(sqrt{frac{2}{3}})^{12}=[(sqrt{frac{2}{3}})^6]^2=(frac{8}{27})^2 = frac{64}{729} [/latex] [latex]c)\ (sqrt[3]{7})^3=7\ (sqrt[3]{7})^4 = (sqrt[3]{7})^3cdot sqrt[3]{7}=7sqrt[3]{7}\ (sqrt[3]{7})^6 = [(sqrt[3]{7})^3]^2=7^2=49\ (sqrt[3]{7})^9 = [(sqrt[3]{7})^3]^3 = 7^3=343\ (sqrt[3]{7})^{12}= [(sqrt[3]{7})^3]^4 = 7^4 = 2401[/latex] [latex]d)\ (sqrt[3]{frac{3}{2}})^3 = frac{3}{2}\ (sqrt[3]{frac{3}{2}})^4 = (sqrt[3]{frac{3}{2}})^3 cdot sqrt[3]{frac{3}{2}}=frac{3}{2}sqrt[3]{frac{3}{2}}=frac{3sqrt[3]{3}}{2sqrt[3]{2}}=frac{3sqrt[3]{12}}{4}\ (sqrt[3]{frac{3}{2}})^6 = [(sqrt[3]{frac{3}{2}})^3]^2 = (frac{3}{2})^2=frac{9}{4}\ (sqrt[3]{frac{3}{2}})^9 = [(sqrt[3]{frac{3}{2}})^3]^3 = (frac{3}{2})^3 = frac{27}{8}\ (sqrt[3]{frac{3}{2}})^{12}=[(sqrt[3]{frac{3}{2}})^3]^4 = (frac{3}{2})^4 = frac{81}{16}[/latex]