Mathematics
If the roots of the equation (c2 - ab)x2 - 2(a2 - bc)x + (b2 - ac) = 0 are real and equal, show that either a = 0 or a3 + b3 + c3 = 3abc.
Quadratic Equations
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Answer
Comparing (c2 - ab)x2 - 2(a2 - bc)x + (b2 - ac) = 0 with ax2 + bx + c = 0 we get,
a = (c2 - ab), b = -2(a2 - bc) and c = (b2 - ac).
Since equations has equal roots,
∴ D = 0
⇒ [-2(a2 - bc)]2 - 4 × (c2 - ab) ×(b2 - ac) = 0
⇒ 4(a2 - bc)2 - 4 × (c2b2 - ac3 - ab3 + a2bc ) = 0
⇒ 4[(a2)2 + (bc)2 - 2 × a2 × bc] - (4c2b2 - 4ac3 - 4ab3 + 4a2bc) = 0
⇒ 4(a4 + b2c2 - 2a2bc) - 4c2b2 + 4ac3 + 4ab3 - 4a2bc = 0
⇒ 4a4 + 4b2c2 - 8a2bc - 4c2b2 + 4ac3 + 4ab3 - 4a2bc = 0
⇒ 4a4 + 4b2c2 - 4c2b2 - 8a2bc - 4a2bc + 4ac3 + 4ab3 = 0
⇒ 4a4 - 12a2bc + 4ac3 + 4ab3 = 0
⇒ 4a(a3 - 3abc + c3 + b3) = 0
⇒ 4a = 0 or (a3 + c3 + b3 - 3abc) = 0 [Using Zero-product rule]
⇒ a = 0 or a3 + c3 + b3 = 3abc
Hence, proved a = 0 or a3 + b3 + c3 = 3abc.
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