Question

The numbers a, b and c satisfy a+b+c = 0 and abc = 78. What is
the value of (a+b)(b+c)(c+a)?​

Answer 1

`(a+b)(b+c)(c+a)=-78`

Explanation:

We are given that the numbers a, b and c

`a+b+c=0` ....(1)

`abc=78` ..... (2)

We have to find the value of (a+b)(b+c)(c+a).

From equation we get

`a+b=-c`

`b+c=-a`

`a+c=-b`

Substitute the values then we get

`(a+b)(b+c)(c+a)=(-c)(-a)(-b)`

`(a+b)(b+c)(c+a)=-abc`

Using equation (2) we get

`(a+b)(b+c)(c+a)=-78`

Hence, the value of
`(a+b)(b+c)(c+a)` is -78.