Question

if a+b+c=6 then find the value of (2-a)cube+(2-b)cube+(2-c)cube-3(2-a)(2-b)(2-c)

Answer 1

Hi ...dear...here is your answer......

let x= 2-a.
y = 2-b
z=2-c
..
adding euqations we get..
x+y+z= 0.....
...now if x + y +z =0 then x(cube)+y(cube)+z(cube)-3xyz=0...
..see picture..!!
hope it helped you!!
Regards Brainly Star Community
#shubhendu

Answer 2

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