If $$a^4+\frac{1}{a^4} =34$$​, then the value of $$a^6+\frac{1}{a^6}$$​ is:

Correct option is B

Given:

​$$a^4 + \frac{1}{a^4} = 34 \\\ \\$$​​

Formula Used:

​$$\left( a^2 + \frac{1}{a^2} \right)^2 = a^4 + \frac{1}{a^4} + 2 \\\ \\$$​​

Solution:

$$\left( a^4 + \frac{1}{a^4} \right)^2 = 34^2 \\\ \\a^8 + 2 + \frac{1}{a^8} = 1156\ \\ \\a^8 + \frac{1}{a^8} = 1154 \\\ \\ \left( a^2 + \frac{1}{a^2} \right)^2 = a^4 + \frac{1}{a^4} + 2 \\\ \\\left( a^2 + \frac{1}{a^2} \right)^2 = 34 + 2 = 36 \\\ \\a^2 + \frac{1}{a^2} = \sqrt{36} = 6 \\\ \\ \left( a^2 + \frac{1}{a^2} \right)\left( a^4 + \frac{1}{a^4} \right) = a^6 + \frac{1}{a^6} + a^2 + \frac{1}{a^2} \\\ \\6 \times 34 = a^6 + \frac{1}{a^6} + 6 \\\ \\204 = a^6 + \frac{1}{a^6} + 6 \\\ \\a^6 + \frac{1}{a^6} = 204 - 6 = \bf 198 \\\ \\$$​​