Question : If $\frac{1}{a}(a^2+1)=3$, then the value of $\frac{a^6+1}{a^3}$ is:
Option 1: 9
Option 2: 18
Option 3: 27
Option 4: 1
Answer (1)
25th Jan, 2024
Correct Answer: 18
Solution :
Given:
$\frac{1}{a}(a^2+1)=3$
⇒ $a+\frac{1}{a}=3$
⇒ $(a+\frac{1}{a})^3=3^3$
⇒ $a^3+\frac{1}{a^3}+3×a×\frac{1}{a}(a+\frac{1}{a})=27$
⇒ $a^3+\frac{1}{a^3}+3×3=27$ [as, $a+\frac{1}{a}=3$]
⇒ $a^3+\frac{1}{a^3}=27-9$
$\therefore \frac{a^6+1}{a^3}=18$
Hence, the correct answer is 18.
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