if a + b + c then the value of (2-a)^3 + (2-b)^3 + (2-c)^3 - 3(2-a)(2-b)(2-c) = ?
Answers
Step-by-step explanation:
a
3
+b
3
+c
3
−3abc=(a+b+c)(a
2
+b
2
+c
2
−ab−bc−ca)
So, if a+b+c=0, a
3
+b
3
+c
3
=3abc
The given expression can be written as
(2a−b)
3
+(b−2c)
3
+(2c−2a)
3
We can see here that 2a−b+b−2c+2c−2a=0.
Hence, following the above concept
(2a−b)
3
+(b−2c)
3
+(2c−2a)
3
=6(2a−b)(b−2c)(c−a)
Hence, options A, B, C are correct
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)}