Question : If $x-\frac{1}{x}=5, x \neq 0$, then what is the value of $\frac{x^6+3 x^3-1}{x^6-8 x^3-1} ?$
Option 1: $\frac{3}{8}$
Option 2: $\frac{13}{12}$
Option 3: $\frac{4}{9}$
Option 4: $\frac{11}{13}$
Correct Answer: $\frac{13}{12}$
Solution :
Given: $x-\frac{1}{x}=5$
Cubing both sides, we get
$(x-\frac{1}{x})^3=5^3$
⇒ $x^3-\frac{1}{x^3}-3(x\times\frac{1}{x})(x-\frac{1}{x})=125$
⇒ $x^3-\frac{1}{x^3}-3\times5=125$
⇒ $x^3-\frac{1}{x^3}=140$
Thus,$\frac{x^6+3 x^3-1}{x^6-8 x^3-1}$
Taking $x^3$ as common, we get:
= $\frac{x^3+3-\frac{1}{x^3}}{x^3-8-\frac{1}{ x^3}}$
= $\frac{x^3-\frac{1}{x^3}+3}{x^3-\frac{1}{ x^3}-8}$
Putting the values, we get:
= $\frac{140+3}{140-8}$
= $\frac{143}{132}$
= $\frac{13}{12}$
Hence, the correct answer is $\frac{13}{12}$.
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