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Question
Solve that `|(x + "a", "b", "c"),("a", x + "b", "c"),("a", "b", x + "c")|` = 0
Solution
`|(x + "a", "b", "c"),("a", x + "b", "c"),("a", "b", x + "c")|` = 0
Put x = 0
`|(0 + "a", "b", "c"),("a", 0 + "b", "c"),("a", "b", 0 + "c")|` = 0
`|("a", "b", "c"),("a", "b", "c"),("a", "b", "c")|` = 0
= 0
x = 0 satisfies the given equation. x = 0 is a root of the given equation
Since three rows are identical. x = 0 is a root of multiplicity 2.
Since the degree of the product of the leading diagonal elements (x + a)(x + b)(x + c) is 3.
There is one more root for the given equation.
Put x = – (a + b + c)
`|((-"a" + "b" + "c") + "a", "b", "c"),("a", -("a" + "b" + "c") + "b", "c"),("a", "b", -("a" + "b" + "c") + "c")|` = 0
`|(-"b" - "c", "b", "c"),("a", -"a" - "c", "c"),("a", "b", -"a" - "b")|` = 0
`"C"_1 -> "C"_1 + "C"_2 + "C"_3`
`|(-"b" - "c" + "b" + "c", "b", "c"),("a" - "a" - "c" + "c", -"a" - "c", "c"),("a" + "b" - "a" - "b", "b", -"a" - "b")|` = 0
`|(0, "b", "c"),(0, -"a" - "c", "c"),(0, "b", -"a" - "b")|` = 0
0 = 0
∴ x = – (a + b + c) satisfies the given equation.
Hence, the required roots of the given equation are x = 0, 0, – (a + b + c)
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