If x³ – 1\over y³ ∝ x³ + 1\over y³, then show that x ∝ 1\over y.
WBBSE
Class X
Mathematics
Variation
03 Mark
2025
(x³ – 1\over y³) ∝ (x³ + 1\over y³)
⇒ x³ + 1\over y³ ∝ x³ – 1\over y³
⇒ x³ + 1\over y³ = k (x³ – 1\over y³)
⇒ x³ + 1\over y³ = k x³ – k 1\over y³
⇒ 1\over y³ + k 1\over y³ = k x³ – x³
⇒ 1\over y³(1 + k) = x³(k – 1)
⇒ x³(k – 1) = 1\over y³(1 + k)
⇒ x³(k – 1) = (1 + k)/(k – 1) 1\over y³
⇒ x³ ∝ 1\over y³
⇒ x ∝ 1\over y (Hence Proved)
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