The value of $(a^{\frac{2}{3}}+2a^{\frac{1}{2}}+3a^{\frac{1}{3}}+2a^{\frac{1}{6}}+1)(a^{\frac{1}{3}}-2a^{\frac{1}{6}}+1)-a^{\frac{1}{2}}(a^{\frac{1}{2}}-2)$ when a = 7 , is :

1

The value of $(a^{\frac{2}{3}}+2a^{\frac{1}{2}}+3a^{\frac{1}{3}}+2a^{\frac{1}{6}}+1)(a^{\frac{1}{3}}-2a^{\frac{1}{6}}+1)-a^{\frac{1}{2}}(a^{\frac{1}{2}}-2)$ when a = 7

Now, Put a^1/6 = k

Then according to the question,

= (k⁴ + 2k³ + 3k² + 2k + 1) (k² - 2k + 1) - k³ (k³ - 2)

= k⁶ + 2k⁵ + 3k⁴ + 2k³ + k² - 2k⁵ - 4k⁴ - 6k³ - 4k² - 2k + k⁴ + 2k³ + 3k² + 2k + 1 - k⁶ + 2k³

= 1