if ` alpha, beta, gamma ` are the roots of the equation ` x^(3) + 3x + 2=0 " then " (alpha^(3) +beta^(3)+gamma^(3))/(alpha^(2) +beta^(2)+gamma^(2))`

Explanation

`alpha , beta , gamma` are the roots of the equation
`x^(3) + 3x + 2=0`
` x^(3) + 3x + 2=0`
`alpha + beta + gamma =0`
` alphabeta + beta gamma + gamma alpha =3`
` alpha beta gamma = -2`
Now ` (alpha^(3) +beta^(3)+gamma^(3))/(alpha^(2)+beta^(2)+gamma^(2)) = ((alpha +beta+gamma) (alpha^(2) +beta^(2) +gamma^(2) -alphabeta -betagamma-gammaalpha)+3alpha betagamma)/((alpha+beta+gamma)^(2) -2(alphabeta+betagamma+gammaalpha))`
`= (3alpha betagamma)/(-2 (alphabeta +betagamma+gammaalpha)) " " (therefore alpha + beta + gamma=0)`
` = (3 xx -2)/(-2.3) =1 `