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If two roots of x^3-a x^2+b x-c=0 are eq...

If two roots of `x^3-a x^2+b x-c=0` are equal in magnitude but opposite in signs, then prove that `a b=c `

Text Solution

Verified by Experts

The correct Answer is:
`x^(3) - 64 = 0`
Let the roots be ` x _(1), - x_(1), x_(2)`. Then
` x_(1) - x_(1) + x_(2) =a `
Hence, x =a is a root of the given equation
`therefore a^(3) - a^(3) + ab - c = 0`
`rArr ab = c `
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