Express large and small. A=3root3,B=4root4,C=6root6 ?

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Answer 1 · 2018-04-20T10:52:49

$A < B < C$ $A < B < C$

Let,

$A = 3 \sqrt{3}, B = 4 \sqrt{4}, C = 6 \sqrt{6}$ $A=3sqrt3,B=4sqrt4, C=6sqrt6$

Squaring we get,

$A^{2} = {(3 \sqrt{3})}^{2} = 9 \times 3 = 27$ $A^2=(3sqrt3)^2=9xx3=27$

$B^{2} = {(4 \sqrt{4})}^{2} = 16 \times 4 = 64$ $B^2=(4sqrt4)^2=16xx4=64$

$C^{2} = {(6 \sqrt{6})}^{2} = 36 \times 6 = 216$ $C^2=(6sqrt6)^2=36xx6=216$

$i . e . A^{2} < B^{2} < C^{2} \Rightarrow A < B < C$ $i.e.A^2 < B^2 < C^2=>A < B < C$

OR

If your question is of type given below ,then see the following
answer.

$A = \sqrt[3]{3}, B = \sqrt[4]{4} and C = \sqrt[6]{6}$ $A=root(3)3, B=root(4)4 and C=root(6)6$

Now, $LCM of : 3 , 4 , 6 is 12$ $"LCM of : 3 , 4 , 6 is 12"$

So,

$A^{12} = {(\sqrt[3]{3})}^{12} = {(3^{\frac{1}{3}})}^{12} = 3^{4} = 81$ $A^12=(root(3)3)^12=(3^(1/3))^12=3^4=81$

$B^{12} = {(\sqrt[4]{4})}^{12} = {(4^{\frac{1}{4}})}^{12} = 4^{3} = 64$ $B^12=(root(4)4)^12=(4^(1/4))^12=4^3=64$

$C^{12} = {(\sqrt[6]{6})}^{12} = {(6^{\frac{1}{6}})}^{12} = 6^{2} = 36$ $C^12=(root(6)6)^12=(6^(1/6))^12=6^2=36$

$i . e . 36 < 64 < 81$ $i.e. 36 < 64 < 81$

$\Rightarrow C^{12} < B^{12} < A^{12}$ $=>C^12 < B^12 < A^12$

$\Rightarrow C < B < A$ $=> C < B < A$