Question

find the values of a, b and c if a+b+c=0, abc=6 and(a+1)(b+1)(c+1)=0​

Answer 1

Answer:

$a + b + c = 0 \\ (a + 1)(b + 1)(c + 1) = 0 \\ ab + b + a + 1(c + 1) = abc + bc + ca + c + ba + b + a + 1 =$

$abc + bc + ba + ac + a + b + c + 1 = 6 + \frac{6}{c} + \frac{6}{a} + \frac{6}{b} + a + b + c + 1$

$6 + \frac{6}{abc} + a + b + c + 1 = 0$

$6 + \frac{6}{6} + a + b + c + 1 = a + b + c = - 8$

your equation has given a+b+c=0

it's wrong because I tried to solve this equation

I got a+b+c=-8

I struggle to solve for this equation