CDS 2026 (I) Elementary Mathematics, question 3

Topic: Algebra

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Q3

If a2b2+c2=b2c2+a2=c2a2+b2\frac{a^{2}}{b^{2}+c^{2}}=\frac{b^{2}}{c^{2}+a^{2}}=\frac{c^{2}}{a^{2}+b^{2}}, then what is the value of a4+b4+c4a^{4}+b^{4}+c^{4} equal to ?

  1. (a) a2b2+b2c2+c2a2a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}Correct answer
  2. (b) 2(a2b2+b2c2+c2a2)2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})
  3. (c) 3(a2b2+b2c2+c2a2)3(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})
  4. (d) 4(a2b2+b2c2+c2a2)4(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})

Solution.

Answer: (a) a2b2+b2c2+c2a2a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}

  1. Put S=a2+b2+c2S = a^{2} + b^{2} + c^{2}.
  2. Equality of a2/(S−a2)a^{2}/\left( S - a^{2} \right) and b2/(S−b2)b^{2}/\left( S - b^{2} \right) gives S(a2−b2)=0S\left( a^{2} - b^{2} \right) = 0.
  3. For real a, b, c with the given fractions defined, S>0S > 0; hence a2=b2=c2a^{2} = b^{2} = c^{2}.
  4. Thus a4+b4+c4a^{4} + b^{4} + c^{4} and a2b2+b2c2+c2a2a^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2} both equal three times the same fourth power.