If the roots ff the equation (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0 are equal, then prove that either a = 0 or a3 + b3 + c3 = 3abc "

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Given equation is: (c 2 – ab) x 2 – 2 (a 2 – bc) x + (b 2 – ac) = 0 To prove: a = 0 or a 3 + b 3 + c 3 = 3abc Proof: From the given equation, we have a = (c2 – ab) b = –2 (a 2 – bc) c = (b 2 – ac) It is being given that equation has real and equal roots ∴...
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Given equation is: (c2–ab)x2– 2 (a2–bc)x+ (b2–ac) = 0 To prove: a= 0 ora3+b3+c3= 3abc Proof: From the given equation, we have a= (c2–ab) b= –2 (a2–bc) c= (b2–ac) It is being given that equation has real and equal roots ∴D= 0 ⇒b2– 4ac= 0 On substituting respective values ofa,bandcin above equation, we get [–2 (a2–bc)]2– 4 (c2–ab) (b2–ac) = 0 4 (a2–bc)2– 4 (c2b2–ac3–ab3+a2bc) = 0 4 (a4+b2c2– 2a2bc) – 4 (c2b2–ac3–ab3+a2bc) = 0 ⇒a4+b2c2– 2a2bc–b2c2+ac3+ab3–a2bc= 0 ⇒a4+ab3+ac3–3a2bc= 0 ⇒a[a3+b3+c3– 3abc] = 0 ⇒a= 0 ora3+b3+c3= 3abc read less
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