Let `f(x)= x^(3) + x + 1` and P(x) be a cubic polynomial such that P(0) = -1 and roots of f(0) = 1 ; P(x) = 0 are the squares of the roots of f(x) = 0 . Then find the value of P(4).

Explanation

According to the question,
`f(x) = x^(3) + x + 1 = (x - alpha ) (x- beta) (x - gamma) " "`…(1)
and `P(x) = k(x - alpha ^(2))(x - beta^(2))(x - gamma^(2))`
Given `P(0) = - 1 rArr - kalpha^(2)beta^(2)gamma^(2) = - 1`
Also `f(0) = 1 rArr - k alpha beta gamma = 1`
`therefore -K(-1)^(2) = - 1 rArr k = 1`
`therefore P(x) = (x - alpha ^(2))(x - beta^(2)) (x - gamma^(2))`
Now , P(4) = `(2 - alpha) (2 - beta) (2 - gamma) (2 + alpha ) (2+beta) (2+gamma)`
`= f (2) (-f(-2))`
`= (2^(3) + 2 + 1 ) (-(-2)^(3) - (-2) - 1)`
`11xx9 = 99`