Question

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

Answer 1

Answer:

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

Step-by-step 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.