B + B/A = 1⇒ (A2 + B2)/AB = 1⇒ (A2 + B2) = ABAs we know,(A3 + B3) = (A + B) (A2 + B2 - AB)⇒ (A3 + B3) = (A + B) (AB - AB)⇒ (A3 + B3) = (A + B) × 0⇒ (A3 + B3) = 0

"> B + B/A = 1⇒ (A2 + B2)/AB = 1⇒ (A2 + B2) = ABAs we know,(A3 + B3) = (A + B) (A2 + B2 - AB)⇒ (A3 + B3) = (A + B) (AB - AB)⇒ (A3 + B3) = (A + B) × 0⇒ (A3 + B3) = 0

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If \(\frac{A}{B}\; + \;\frac{B}{A}\; = \;1,\) then find A3 + B3.

General Aptitude Algebra in General Aptitude . 2 months ago

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A/B + B/A = 1⇒ (A2 + B2)/AB = 1⇒ (A2 + B2) = ABAs we know,(A3 + B3) = (A + B) (A2 + B2 - AB)⇒ (A3 + B3) = (A + B) (AB - AB)⇒ (A3 + B3) = (A + B) × 0⇒ (A3 + B3) = 0

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