Question

if a+b+c=6;bc+ca+ab=11;abc=6;then the value of (1-a)(1-b)(1-c)​

Answer 1

Answer:

0 ans

Step-by-step explanation:

(1-a)(1-b)(1-c)= (1-b-a+ab) (1-c)

= 1-b-a+ab-c+bc+ac-abc

=1-(a+b+C) +(ab+bc+ac)-(abc)

=1-6+11-6

=12-12

=0 ans

Answer 2

The value of the expression is 0.

Given: The given values are a+b+c=6;bc+ca+ab=11;abc=6.

To find: We have to find the value of (1-a)(1-b)(1-c).

Solution:

The expression is (1-a)(1-b)(1-c).

Multiplying (1-a) and (1-b) we get-

$(1 - a)(1 - b) = 1 - a - b + ab$

Now multiplying the result with (1-c) we get-

$(1 - a - b + ab)(1 - c) \\ 1 - a - b + ab - c + ac + b - abc$

The overall expression can be written as-

$1 - (a + b + c) + (ab + ac + bc) - abc$

Thus putting the value of (a+b+c) which is 6, bc+ca+ab which is 11 and abc which is 6 in the expression we get-

$1 - 6 + 11 - 6 \\ = 0$

Thus, the value of the expression is 0.