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यदि समीकरण (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0 के मूल बराबर हैं तो सिद्ध कीजिए कि a = 0 या a3 + b3 + c3 = 3abc |
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Answer» समीकरण (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0 A = c2 – ab, B = -2(a2 – bc), C = b2 – ac ∵ समीकरण के मूल बराबर हैं। ∴ (B)2 – 4AC = 0 [- 2(a2 – bc)]2 – 4(c2 – ab)(b2 – ac) = 0 4(a2 – bc)2 – 4(b2c2 – ac3 – ab3 + a2bc) = 0 4(a4 + b2c2 – 2a2bc – b2c2 + ac3 + ab3 – a2bc) = 0 a4 + ab3 + ac3 – 3a2bc = 0 a(a3 + b3 + c3 – 3abc) = 0 a = 0 तथा a3 + b3 + c3 – 3abc = 0 a3 + b3 + c3 = 3abc अतः a = 0 या a3 + b3 + c3 = 3abc इति सिद्धम्। |
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