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Usingproperties of determinants, prove that following:`|"a"+"b"+2"c""a""b""c""b"+"c"+2"a""b""c""a""c"+"a"+2"b"|=2("a"+"b"+"c")^3`

Answer» Let the value of the given determinant be `Delta`. Then,
`Delta = |{:(a+b+2c, " "a, " "b),(" "c, b+c+2a, " "b),(" "c, " "a, c+a+2b):}|`
`= |{:(a+b+c, -(a+b+c), " "0),(" "c, b+c+2a, " "b),(" "0, -(a+b+c), (a+b+c)):}| {{:("by"R_(1) to (R_(1) -R_(2))),("and"R_(3) to (R_(3) - R_(2))):}}`
`=(a+b+c)^(2) * |{:(1, " "-1, " "0),(c, b+c+2a, " "b),(0, " "-1, " "1):}|`
`["taking(a+b+c) common from each one of "R_(1) "and"R_(3)]`
`=(a+b+c)^(2) * |{:(1, " "-1, " "0),(0, b+2c+2a, " "b),(0, " "-1, " "1):}| " "[R_(2) to R_(2) -cR_(1)]`
`=(a+b+c)^(2) *1* |{:(b+2c+2a, b),(-1, 1):}|["expanded by"C_(1)]`
`=(a+b+c)^(2) * (b+2c+2a+b) = 2(a+b+c)^(3)`
Hence, `Delta =2(a+b+c)^(3)`


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