Answer

Given cubic polynomial

p(x)=x³+3x²-x-3

Comparing the given polynomial with

ax³ + bx² + cx + d,we get a=1,b=3,c= -1,d= -3

Further given zeros are 1,-1 and -3

p(1)=(1)³ + 3(1)² - 1 - 3 =

p(-1)=(-1)³ + 3(-1)² - (-1) - 3 =

p(-3)=(-3)³ + 3(-3)² - (-3) - 3 =

Therefore 1,-1 and -3 are the zeros of x³+3x²-x-3

So,we take α=1, β=-1 and γ= -3
Now,α+β+γ=1+(-1)+(-3)=-

αβ+βγ+ γα=1(-1)+(-1)(-3)+(-3)1

=-1+3-3=-