Question

please answer question no 101​

Answer 1

a+b+c=6

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

Then (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