Arrange the following in increasing order:

A) $$\sqrt[4]{3}$$
B) $$\sqrt[3]{2}$$
C) $$\sqrt[6]{5}$$
D) $$\sqrt[6]{7}$$

Choose the correct answer from the options given below :

We can rewrite the following as:

A) $$\sqrt[4]{3}$$ = $$3^{\dfrac{1}{4}}$$

B) $$\sqrt[3]{2}$$ = $$2^{\dfrac{1}{3}}$$

C) $$\sqrt[6]{5}$$ = $$5^{\dfrac{1}{6}}$$

D) $$\sqrt[6]{7}$$ = $$7^{\dfrac{1}{6}}$$

Now, we will make the denominator of all equal:

A)$$3^{\dfrac{3}{12}}$$

B) $$2^{\dfrac{4}{12}}$$

C) $$5^{\dfrac{2}{12}}$$

D) $$7^{\dfrac{2}{12}}$$

Simplifying further, we get, 

A)$$3^{\dfrac{3}{12}}$$ = $$\left(3^3\right)^{\dfrac{1}{12}}=\left(27\right)^{\dfrac{1}{12}}$$

B) $$2^{\dfrac{4}{12}}$$ = $$\left(2^4\right)^{\dfrac{1}{12}}=\left(16\right)^{\dfrac{1}{12}}$$

C) $$5^{\dfrac{2}{12}}$$ = $$\left(5^2\right)^{\dfrac{1}{12}}=\left(25\right)^{\dfrac{1}{12}}$$

D) $$7^{\dfrac{2}{12}}$$ = $$\left(7^2\right)^{\dfrac{1}{12}}=\left(49\right)^{\dfrac{1}{12}}$$

Arranging them in sequence: We get: B < C < A < D