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Question: If \(x^{3} + x + 1 = 0\) is a root of the equation \(\alpha^{3}\beta^{3}\gamma^{3}\), where p and q ...

If x3+x+1=0x^{3} + x + 1 = 0 is a root of the equation α3β3γ3\alpha^{3}\beta^{3}\gamma^{3}, where p and q are real, then x4−2x3+x=380x^{4} - 2x^{3} + x = 380=

A

5,−4,1±5−325, - 4,\frac{1 \pm 5\sqrt{- 3}}{2}

B

−5,4,−1±5−32- 5,4, - \frac{1 \pm 5\sqrt{-}3}{2}

C

(4, 7)

D

5,4,−1±5−325,4,\frac{- 1 \pm 5\sqrt{-}3}{2}

Answer

5,−4,1±5−325, - 4,\frac{1 \pm 5\sqrt{- 3}}{2}

Explanation

Solution

Since x2+ax+b=0x^{2} + ax + b = 0 is a root, therefore x2+bx+a=0x^{2} + bx + a = 0 will be other root. Now sum of the roots a≠ba \neq b and product of roots a+b+4=0a + b + 4 = 0. Hence a+b−4=0a + b - 4 = 0.