CDS 2026 (I) Elementary Mathematics, question 84

Topic: Algebra

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Passage · Q83 to Q84

16(x4+1x4)−257=016\left( x^{4}+\frac{1}{x^{4}} \right)-257=0

Q84

What is (x3+1x3)\left( x^{3}+\frac{1}{x^{3}} \right) equal to ?

  1. (a) 658\frac{65}{8}Correct answer
  2. (b) 638\frac{63}{8}
  3. (c) 618\frac{61}{8}
  4. (d) 598\frac{59}{8}

Solution.

Answer: (a) 658\frac{65}{8}

  1. Divide by 16: x4+1x4=25716x^{4} + \frac{1}{x^{4}} = \frac{257}{16}, so (x2+1x2)2=28916\left( x^{2} + \frac{1}{x^{2}} \right)^{2} = \frac{289}{16} and x2+1x2=174x^{2} + \frac{1}{x^{2}} = \frac{17}{4}.
  2. Then (x+1x)2=174+2=254\left( x + \frac{1}{x} \right)^{2} = \frac{17}{4} + 2 = \frac{25}{4}, so s=x+1x=52s = x + \frac{1}{x} = \frac{5}{2}, taking the positive root.
  3. x3+1x3=s3−3s=1258−152=658x^{3} + \frac{1}{x^{3}} = s^{3} - 3s = \frac{125}{8} - \frac{15}{2} = \frac{65}{8}.

Trap: x=−2x = -2 also satisfies the equation and gives −658-\frac{65}{8}; the question does not fix the sign, and only the positive value is among the options.