CAT 2025 Slot 3 · Quant · Question 4

Squaring x+\frac{1}{x} gives {x}^{2}+2+\frac{1}{{x}^{2}}, which is 25+2=27.

So x+\frac{1}{x}=\sqrt{27}=3\sqrt{3}, taking the positive root since x>0.

Cubing, {x}^{3}+\frac{1}{{x}^{3}}={\left( x+\frac{1}{x}\right)}^{3}-3\left( x+\frac{1}{x}\right).

That is {\left( 3\sqrt{3}\right)}^{3}-3\left( 3\sqrt{3}\right)=81\sqrt{3}-9\sqrt{3}=72\sqrt{3}.

Squaring the given expression, {x}^{4}+\frac{1}{{x}^{4}}={\left({x}^{2}+\frac{1}{{x}^{2}}\right)}^{2}-2=625-2=623.

Now multiply the fourth and third powers: \left({x}^{4}+\frac{1}{{x}^{4}}\right)\left({x}^{3}+\frac{1}{{x}^{3}}\right)={x}^{7}+\frac{1}{{x}^{7}}+x+\frac{1}{x}.

The left side is 623\times 72\sqrt{3}=44856\sqrt{3}.

Subtracting the stray x+\frac{1}{x}=3\sqrt{3}, {x}^{7}+\frac{1}{{x}^{7}}=44856\sqrt{3}-3\sqrt{3}=44853\sqrt{3}.

Shortcut. To reach power 7, multiply the powers that add to it and subtract the leftover: \left({x}^{4}+\frac{1}{{x}^{4}}\right)\left({x}^{3}+\frac{1}{{x}^{3}}\right) leaves exactly {x}^{7}+\frac{1}{{x}^{7}} plus {x}^{1}+\frac{1}{{x}^{1}}, because 4-3=1. The same trick (m+n and m-n) reaches any power quickly.