Question

If a + b + c = 6, find value of
(2-a)^3 + (2 - b)^3 + (2 - 0)^3 - 3(2-a)(2 - b)(2 -c)​

Answer 1

Answer:

Step-by-step explanation:

a+b+c=6

Therefore (2-a)+(2-b)+(2-c)=0

Now (2-a)^3+(2-b)^3+(2-c)^3–3(2-a)(2-b)(2-c)

={(2-a)+(2-b)+(2-c)}{(2-a)^2+(2-b)^2+(2-c)^2-(2-a)(2-b)-(2-b)(2-c)-(2-c)(2-a)} (since a^3+b^3+c^3–3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)

={2-a+2-b+2-c}{(2-a)^2+(2-b)^2+(2-c)^2–(2-a)(2-b)-(2-b)(2-c)-(2-c)(2-a)}

={6-(a+b+c)}{(2-a)^2+(2-b)^2+(2-c)^2-(2-a)(2-b)-(2-b)(2-c)-(2-c)(2-a)}

={6–6}{(2-a)^2+(2-b)^2+(2-c)^2–(2-a)(2-b)-(2-b)(2-c)-(2-c)(2-a)}

=0×{(2-a)^2+(2-b)^2+(2-c)^2-(2-a)(2-b)-(2-b)(2-c)-(2-c)(2-a)}

=0

Answer 2

Answer:

3[1-2(ab+bc+ca)+abc]

hope it helps