CAT 2023 Slot 3 · Quant · Question 4

Everything here is built from u=x+\frac{1}{x}, so work down to that first.

Squaring, {x}^{2}+\frac{1}{{x}^{2}}={u}^{2}-2.

Squaring again, {x}^{4}+\frac{1}{{x}^{4}}={\left({u}^{2}-2 \right)}^{2}-2.

Once more, {x}^{8}+\frac{1}{{x}^{8}}={\left({\left({u}^{2}-2 \right)}^{2}-2 \right)}^{2}-2.

Setting this to 47, {\left({\left({u}^{2}-2 \right)}^{2}-2 \right)}^{2}=49.

Since x is positive, every one of these sums is positive, so take the positive root: {\left({u}^{2}-2 \right)}^{2}-2=7.

Then {\left({u}^{2}-2 \right)}^{2}=9, so {u}^{2}-2=3 and {u}^{2}=5, giving u=\sqrt{5}.

For the cube, {x}^{3}+\frac{1}{{x}^{3}}={u}^{3}-3u=5\sqrt{5}-3\sqrt{5}=2\sqrt{5}.

Now apply the same cube identity one level up, with v={x}^{3}+\frac{1}{{x}^{3}}: {x}^{9}+\frac{1}{{x}^{9}}={v}^{3}-3v.

Substituting v=2\sqrt{5}, {v}^{3}=8\times 5\sqrt{5}=40\sqrt{5} and 3v=6\sqrt{5}.

So {x}^{9}+\frac{1}{{x}^{9}}=40\sqrt{5}-6\sqrt{5}=34\sqrt{5} — the fourth option.

Shortcut. Climbing from the 8th power down to u is three square roots, each of which lands on a whole number — a sign the path is right. Going back up, note that the identity {a}^{3}+\frac{1}{{a}^{3}} applies to {x}^{3} just as it does to x, so the ninth power needs no new work.