Let's expand the determinant. The general formula for a 3x3 determinant \(\left| {\begin{array}{*{20}{c}} p&q&r\\ s&t&u\\ v&w&x \end{array}} \right|\) is \(p(tx - uw) - q(sx - uv) + r(sw - tv)\).
Applying this to our determinant:
\(\left| {\begin{array}{*{20}{c}} {{a}}&{{b}}&{{c}}\\ {{b}}&{{c}}&{{a}}\\ {{c}}&{{a}}&{{b}} \end{array}} \right| = a(c \cdot b - a \cdot a) - b(b \cdot b - c \cdot a) + c(b \cdot a - c \cdot c)\)
\(= a(bc - a^2) - b(b^2 - ac) + c(ab - c^2)\)
\(= abc - a^3 - b^3 + abc + abc - c^3\)
\(= 3abc - (a^3 + b^3 + c^3)\)
Using Algebraic Identities
We need to find the value of \(3abc - (a^3 + b^3 + c^3)\). We can use a standard algebraic identity that relates \(a+b+c\), \(ab+bc+ca\), \(a^2+b^2+c^2\), \(a^3+b^3+c^3\), and \(abc\).
The identity is:
\(a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab - bc - ca)\)
From the given conditions, we know \(a+b+c = 4\) and \(ab+bc+ca = 0\).
We need to find the value of \(a^2+b^2+c^2\). We can get this from the identity:
\((a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca)\)
Substitute the given values:
\((4)^2 = a^2 + b^2 + c^2 + 2(0)\)
\(16 = a^2 + b^2 + c^2 + 0\)
So, \(a^2 + b^2 + c^2 = 16\).
Now, substitute the known values into the main algebraic identity:
\(a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2 - ab - bc - ca)\)
Additional Information: Properties of Determinants and Algebraic Expressions
Determinants are scalar values associated with square matrices. They have many applications in linear algebra, such as solving systems of linear equations, finding matrix inverses, and calculating areas/volumes.
The specific determinant in this problem is a type of cyclic determinant. These often have symmetric or patterned expansions that can be simplified using algebraic identities.
The algebraic identity \(a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab - bc - ca)\) is very useful. A special case arises when \(a+b+c = 0\). In this case, the identity becomes \(a^3 + b^3 + c^3 - 3abc = (0)(a^2+b^2+c^2 - ab - bc - ca) = 0\), which simplifies to \(a^3 + b^3 + c^3 = 3abc\). This special case was not directly applicable here as \(a+b+c = 4 \neq 0\).
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