Let `alpha,beta,gamma` be the roots of the equation `8x^3 +1001x+ 2008=0` then the value `(alpha+beta)^3+(beta+gamma)^3+(gamma+alpha)^3` is

Options

A. 251
B. 751
C. 735
D. 753

Answer

D

Explanation

We have `alpha+beta+gamma = 0, alpha beta gamma = - (2008)/(8) = - 251`
Let `A = alpha + beta, B = Beta + gamma, C = gamma + alpha`. Then, `A + B + C = 2(alpha + beta + gamma) = 0`
`therefore" "A^(3) + B^(3) + C^(3) = 3ABC`
`rArr" "(alpha+beta)^(3)+(beta+gamma)^(3)+(gamma+alpha)^(3) = 3(alpha+beta)(beta+gamma)(gamma+alpha)" "=gamma alpha beta" "[therefore" "alpha+beta+gamma=0]`
`" "-3 xx -251 = 753`