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 :
Answer & explanation
Correct answer: option 3
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 a1/6 = 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