Помогите! Вычислите:

Решите задачу:

$( \sqrt{3} )^2=3$
$\sqrt[3]{8^2} = \sqrt[3]{8*8}=2*2=4$
$\sqrt[3]{125^2} = \sqrt[3]{(5^3)^2} = \sqrt[3]{5^6}=5^2=25$
$\sqrt[4]{81^3}= \sqrt[4]{(3^4)^3}= \sqrt[4]{3^{12}}=3^3= 27$
$\sqrt{49^3}= \sqrt{(7^2)^3} = \sqrt{7^6}=7^3=343$
$\sqrt[3]{27^2}= \sqrt[3]{(3^3)^2}= \sqrt[3]{3^6} =3^2=9$
$\sqrt[4]{16^3}= \sqrt[4]{(2^4)^3} = \sqrt[4]{2^{12}} =2^3=8$
$\sqrt[5]{32^4}= \sqrt[5]{(2^5)^4}= \sqrt[5]{2^{20}}= 2^4=16$

$\sqrt[4]{9^2}= \sqrt[4]{(3^2)^2}= \sqrt[4]{3^4} =3$
$\sqrt[4]{25^2}= \sqrt[4]{(5^2)^2}= \sqrt[4]{5^4}=5$
$\sqrt[6]{8^2}= \sqrt[6]{(2^3)^2}= \sqrt[6]{2^6} =2$
$\sqrt[6]{16^3}= \sqrt[6]{(4^2)^3}= \sqrt[6]{4^{6}}=4$
$\sqrt[6]{27^2}= \sqrt[6]{(3^3)^2}= \sqrt[6]{3^6}=3$
$\sqrt[6]{81^3}= \sqrt[6]{(3^4)^3}= \sqrt[6]{3^{12}} =3^2=9$
$\sqrt[200]{49^{100}} = \sqrt[200]{(7^2)^{100}} = \sqrt[200]{7^{200}} =7$
$\sqrt[300]{125^{100}} = \sqrt[300]{(5^3)^{100}} = \sqrt[300]{5^{300}}=5$

$\sqrt[4]{81} = \sqrt[4]{3^4} =3$
$\sqrt[4]{625}= \sqrt[4]{5^4}= 5$
$\sqrt[4]{160 000} = \sqrt[4]{16*10000} = \sqrt[4]{2^4*10^4} = 2*10=20$
$\sqrt[4]{0.0625}= \sqrt[4]{625*0.0001} = \sqrt[4]{5^4*0.1^4} =5*0.1=0.5$
$\sqrt[6]{729}= \sqrt[6]{3^6}=3$
$\sqrt[6]{64000000}= \sqrt[6]{64*100000}= \sqrt[6]{2^6*10^6}= 2*10=20$
$\sqrt[6]{0.000729}= \sqrt[6]{729*0.000001}= \sqrt[6]{3^6*0.1^6}=3*0.1=0.3$