Which of the following statement(s) is/are TR | vedclass

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(A) To compare radicals,we convert them to a common root (index) by taking the Least Common Multiple $(LCM)$ of the indices.$I.$ Indices are $2, 3, 4$

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Only $I$ and $II$

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(A) To compare radicals,we convert them to a common root (index) by taking the Least Common Multiple $(LCM)$ of the indices.$I.$ Indices are $2, 3, 4$. $LCM(2, 3, 4) = 12$.$\sqrt{12} = 12^{1/2} = 12^{6/12} = (12^6)^{1/12} = (2,985,984)^{1/12}$$\sqrt[3]{16} = 16^{1/3} = 16^{4/12} = (16^4)^{1/12} = (65,536)^{1/12}$$\sqrt[4]{24} = 24^{1/4} = 24^{3/12} = (24^3)^{1/12} = (13,824)^{1/12}$Since $2,985,984 > 65,536 > 13,824$,statement $I$ is $TRUE$.$II.$ Indices are $3, 4, 6$. $LCM(3, 4, 6) = 12$.$\sqrt[3]{25} = 25^{4/12} = (25^4)^{1/12} = (390,625)^{1/12}$$\sqrt[4]{32} = 32^{3/12} = (32^3)^{1/12} = (32,768)^{1/12}$$\sqrt[6]{48} = 48^{2/12} = (48^2)^{1/12} = (2,304)^{1/12}$Since $390,625 > 32,768 > 2,304$,statement $II$ is $TRUE$.$III.$ Indices are $4, 3, 6$. $LCM(4, 3, 6) = 12$.$\sqrt[4]{9} = 9^{3/12} = (9^3)^{1/12} = (729)^{1/12}$$\sqrt[3]{15} = 15^{4/12} = (15^4)^{1/12} = (50,625)^{1/12}$$\sqrt[6]{24} = 24^{2/12} = (24^2)^{1/12} = (576)^{1/12}$Since $50,625 > 729 > 576$,the order should be $\sqrt[3]{15} > \sqrt[4]{9} > \sqrt[6]{24}$. Thus,statement $III$ is $FALSE$.Therefore,only $I$ and $II$ are true.

Which of the following statement$(s)$ is/are $TRUE$?
$I.$ $\sqrt{12} > \sqrt[3]{16} > \sqrt[4]{24}$
$II.$ $\sqrt[3]{25} > \sqrt[4]{32} > \sqrt[6]{48}$
$III.$ $\sqrt[4]{9} > \sqrt[3]{15} > \sqrt[6]{24}$

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Which of the following statement$(s)$ is/are $TRUE$?$I.$ $\sqrt{12} > \sqrt[3]{16} > \sqrt[4]{24}$$II.$ $\sqrt[3]{25} > \sqrt[4]{32} > \sqrt[6]{48}$$III.$ $\sqrt[4]{9} > \sqrt[3]{15} > \sqrt[6]{24}$