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Question
If the roots of the equation (c2 – ab) x2 – 2 (a2 – bc) x + b2 – ac = 0 in x are equal, then show that either a = 0 or a3 + b3 + c3 = 3abc
Solution
The given quadric equation is `(c^2 - ab)x^2 - 2(a^2 - bc)x + (b^2 - ac) = 0`
THen prove that either a = 0 or `a^3 + b^3 + c^3 = 3abc`
Here
`a = (c^2 - ab), b = -2(a^2 - bc) " and " c = (b^2 - ac)`
As we know that `D = b^2 - 4ac`
Putting the value of `a = (c^2 - ab), b = -2(a^2 - bc) and c = (b^2 - ac)`
The given equation will have real roots, if D = 0
`4a(a^3 + b^3 + c^3 - 3abc) = 0`
`a(a^3 + b^3 + c^3 - 3abc) = 0`
So , either
a = 0
or
`(a^3 + b^3 + c^3 - 3abc) = 0`
`a^3 + b^3 + c^3 = 3abc`
Hence, a = 0 or `a^3 + b^3 + c^3 = 3abc`
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