(2a-b-c)3 + (2b-c-a)3 + (2c-a-b)3 i
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Ans:- Let(2a-b-c) = x, (2b-c-a) = y, (2c-a-b) = z
Now,
=> x³ + y³ + z³ = (x + y + z)³ - 3(x+y)(y+z)(z+x)
=> (2a-b-c)³+(2b-c-a)³+(2c-a-b)³ = (2a-b-c+2b-c-a+2c-a-b)³-3(2a-b-c+2b-c-a)(2b-c-a+2c-a-b)(2c-a-b+2a-b-c)
=> (2a-b-c)³+(2b-c-a)³+(2c-a-b)³ = (0)³ -3(a+b-2c)(b+c-2a)(c+a-2b)
=> (2a-b-c)³+(2b-c-a)³+(2c-a-b)³ = -3(a+b-2c)(b+c-2a)(c+a-2b)
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