4. If a + b + c = 0 then find the value of (a+b-2c)3 +(a+c-2b)3 +(6+c-2a)3 (a+b-20)(a+c-2b)(b +c-2a)
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Answer:
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)
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