CAT 2023 Slot 3 QA Question 2

First, I recall that expressions like $x^n + \frac{1}{x^n}$ can often be related to each other using certain algebraic identities. For example, if I know $x + \frac{1}{x} , I can find$x^2 + $\frac{1}{x^2}$$ by squaring the original expression. Maybe I can use a similar approach here, but since the exponents are 8 and 9, it might involve multiple steps.

Let $y = x + \frac{1}{x}$. I know that $y^2 = x^2 + 2 + \frac{1}{x^2} , so$x^2 + $\frac{1}{x^2}$ = y^2 - 2$. Similarly,$x^3 + $\frac{1}{x^3}$$can be expressed in terms of y as well. But wait, I have$x^8 + $\frac{1}{x^8}$ , which is a much higher power. Maybe I can find a pattern or a recursive formula.

Alternatively, perhaps I can express $x^8 + \frac{1}{x^8} in terms of lower powers. Let me think:$(x^4 + $\frac{1}{x^4}$)^2 = x^8 + 2 + $\frac{1}{x^8}$ , so $x^8 + \frac{1}{x^8} = (x^4 + \frac{1}{x^4})^2 - 2$. Given that $x^8 + \frac{1}{x^8} = 47 , this implies$(x^4 + $\frac{1}{x^4}$)^2 = 49 , so $x^4 + \frac{1}{x^4} = 7$ since x is positive, so we take the positive root.

Now, moving down, $x^4 + \frac{1}{x^4} = 7$. Let me find $x^2 + \frac{1}{x^2}$. Using the same identity: $(x^2 + \frac{1}{x^2})^2 = x^4 + 2 + \frac{1}{x^4}$. So, $(x^2 + \frac{1}{x^2})^2 = 7 + 2 = 9 , which means$x^2 + $\frac{1}{x^2}$ = 3$.

Next, I can find $x + \frac{1}{x}$. Let $y = x + \frac{1}{x}$. Then, $y^2 = x^2 + 2 + \frac{1}{x^2}$. We know $x^2 + \frac{1}{x^2} = 3 , so$y^2 = 3 + 2 = 5 , which means $y = \sqrt{5}$.

Now, We find $x^9 + \frac{1}{x^9}$. Hmm, how can I express this in terms of lower powers? $x^n + \frac{1}{x^n}$ can be expressed using $x^{n-1} + \frac{1}{x^{n-1}}$ and $x + \frac{1}{x}$. Specifically, there's a recursive formula: $x^n + \frac{1}{x^n} = (x + \frac{1}{x})(x^{n-1} + \frac{1}{x^{n-1}}) - (x^{n-2} + \frac{1}{x^{n-2}})$.

Let me apply this for n=9. So, $x^9 + \frac{1}{x^9} = (x + \frac{1}{x})(x^8 + \frac{1}{x^8}) - (x^7 + \frac{1}{x^7})$. But wait, I don't know $x^7 + \frac{1}{x^7}$. Maybe I can find that first.

Similarly, $x^7 + \frac{1}{x^7} = (x + \frac{1}{x})(x^6 + \frac{1}{x^6}) - (x^5 + \frac{1}{x^5})$. Hmm, this seems like it's going to take a lot of steps. Maybe there's a smarter way.

Alternatively, perhaps I can find $x^3 + \frac{1}{x^3}$ first and then build up to 9. Let's try that.

We have $y = x + \frac{1}{x} = \sqrt{5}$. Then, $x^3 + \frac{1}{x^3} = y^3 - 3y$. Compute that: $(\sqrt{5})^3 - 3\sqrt{5} = 5\sqrt{5} - 3\sqrt{5} = 2\sqrt{5}$.

Next, $x^4 + \frac{1}{x^4} = 7 , which we already found. Now,$x^5 + $\frac{1}{x^5}$$can be found using the recursive formula:$x^5 + $\frac{1}{x^5}$ = (x + $\frac{1}{x}$)(x^4 + $\frac{1}{x^4}$) - (x^3 + $\frac{1}{x^3}$)$. Plugging in the values:$\ × 7 - 2\ = 7\ - 2\ = 5\$.

Similarly, $x^6 + \frac{1}{x^6} = (x + \frac{1}{x})(x^5 + \frac{1}{x^5}) - (x^4 + \frac{1}{x^4})$. That would be \ × 5\ - 7 = 5 × 5 - 7 = 25 - 7 = 18$.

Moving on, \( x^7 + $\frac{1}{x^7}$ =