Question

find the value of (2-a)³+(2-b)³+(2-c)³-3(2-a)(2-b)(2-c) where a+b+c=6​

Answer 1

Answer:

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

Step-by-step explanation:

Answer 2

Answer:

So (2-a)³+(2-b)³+(2-c)³-3(2-a)(2-b)(2-c)=0

Step-by-step explanation:

a+b+c=6

(2-a)³+(2-b)³+(2-c)³-3(2-a)(2-b)(2-c)

Now let 2-a= x, 2-b = y and 2-c = z

x+y+c=2-a+2-b+2-c=6-(a+b+c)=6-6=0

So x+y+z=0

Then x³+y³+z³-3xyz=0

So (2-a)³+(2-b)³+(2-c)³-3(2-a)(2-b)(2-c)=0