Question

(2-a)3+(2-b)3+(2-c)3-3(2- a)(2-b)(2-c)when a+ b+ c=6

Answer 1

if x + y +z =0 then x(cube)+y(cube)+z(cube)-3xyz=0
therefore, if x = 2-a ; y = 2-b ; z = 2-c
then x+y+z i.e., 2-a+2-b+2-c=0.this implies that a+b+c=6
then x(cube)+y(cube)+z(cube)-3xyz=0 i.e,
(2-a)3+(2-b)3+(2-c)3-3(2- a)(2-b)(2-c) = 0

Answer 2

Answer:

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