यदि `(2a-1)^(2)+(4b-3)^(2)+(4c+5)^(2)=0` तो `(a^(3)+b^(3)+c^(3)-3abc)/(a^(2)+b^(2)+c^(2))` का मान क्या होगा?

Options

A. `3 3/8`
B. `2 3/8`
C. 0
D. `1 3/8`

Answer

C

Explanation

`(2a-1)^(2)+(4b-3)^(2)+(4c+5)^(2)=0`
to find
`implies (a^(3)+b^(3)+c^(3)-3abc)/(a^(2)+b^(2)+c^(2))`
`implies (1/2(a+b+c)[(a-b)^(2)+(b-c)^(2)+(c-a)^(2)])/(a^(2)+b^(2)+c^(2))`
`implies` then `2a-1=0`
`a=1//2`
`implies 4b-3=0`
`implies b=3/4`
`implies 4c+5=0`
`implies c=(-5)/4`
`=a+b+c=1/2+3/4-5/4=0`
`implies (1/2(a+b+c)[(a-b)^(2)+(b-c)^(2)+(c-a)^(2)])/(a^(2)+b^(2)+c^(2))`
`=(1/2(0)[(a-b)^(2)+(b-c)^(2)+(c-a)^(2)])/(a^(2)+b^(2)+c^(2))=0`