Key Concepts and Formulas

• Determinant Properties:
  • Additivity of a Column/Row: A determinant can be split into the sum of two determinants if a column (or row) consists of the sum of two terms. For example, $\left| {\begin{matrix} a & b & c+d \\ e & f & g+h \\ i & j & k+l \\ \end{matrix} } \right| = \left| {\begin{matrix} a & b & c \\ e & f & g \\ i & j & k \\ \end{matrix} } \right| + \left| {\begin{matrix} a & b & d \\ e & f & h \\ i & j & l \\ \end{matrix} } \right|$.
  • Factoring a Column/Row: A common factor in any single column or row can be factored out of the determinant.
  • Column/Row Swaps: Swapping two columns or two rows of a determinant multiplies its value by $-1$.
• Vandermonde Determinant: The determinant $\left| {\begin{matrix} 1 & x & {{x^2}} \\ 1 & y & {{y^2}} \\ 1 & z & {{z^2}} \\ \end{matrix} } \right|$ is equal to $(y-x)(z-x)(z-y)$. This determinant is non-zero if and only if $x, y, z$ are distinct.
• Scalar Triple Product and Non-Coplanar Vectors: Three vectors $\vec{u} = (u_1, u_2, u_3)$, $\vec{v} = (v_1, v_2, v_3)$, and $\vec{w} = (w_1, w_2, w_3)$ are non-coplanar if and only if their scalar triple product, given by the determinant of the matrix formed by their components, is non-zero: $[\vec{u} \ \vec{v} \ \vec{w}] = \left| {\begin{matrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \\ \end{matrix} } \right| \neq 0$.

Step-by-Step Solution

We are given the determinant equation: $\left| {\begin{matrix} a & {{a^2}} & {1 + {a^3}} \\ b & {{b^2}} & {1 + {b^3}} \\ c & {{c^2}} & {1 + {c^3}} \\ \end{matrix} } \right| = 0$

Step 1: Split the determinant based on the sum in the third column. The third column consists of terms of the form $1 + x^3$. We can use the additivity property of determinants to split this into two separate determinants.

• Why this step? This allows us to isolate the constant part (1) and the cubic part ($a^3, b^3, c^3$) into different determinants, which can then be manipulated independently. $\left| {\begin{matrix} a & {{a^2}} & 1 \\ b & {{b^2}} & 1 \\ c & {{c^2}} & 1 \\ \end{matrix} } \right| + \left| {\begin{matrix} a & {{a^2}} & {{a^3}} \\ b & {{b^2}} & {{b^3}} \\ c & {{c^2}} & {{c^3}} \\ \end{matrix} } \right| = 0 \quad \text{... (1)}$

Step 2: Factor out common terms from the second determinant. In the second determinant, we can factor out $a$ from the first row, $b$ from the second row, and $c$ from the third row.

• Why this step? Factoring simplifies the elements of the determinant and reveals a structure that can be related to the Vandermonde determinant. $\left| {\begin{matrix} a & {{a^2}} & {{a^3}} \\ b & {{b^2}} & {{b^3}} \\ c & {{c^2}} & {{c^3}} \\ \end{matrix} } \right| = abc \left| {\begin{matrix} 1 & a & {{a^2}} \\ 1 & b & {{b^2}} \\ 1 & c & {{c^2}} \\ \end{matrix} } \right|$ Substituting this back into equation (1), we get: $\left| {\begin{matrix} a & {{a^2}} & 1 \\ b & {{b^2}} & 1 \\ c & {{c^2}} & 1 \\ \end{matrix} } \right| + abc \left| {\begin{matrix} 1 & a & {{a^2}} \\ 1 & b & {{b^2}} \\ 1 & c & {{c^2}} \\ \end{matrix} } \right| = 0 \quad \text{... (2)}$

Step 3: Transform the first determinant to match the standard Vandermonde form. Let $D_1 = \left| {\begin{matrix} a & {{a^2}} & 1 \\ b & {{b^2}} & 1 \\ c & {{c^2}} & 1 \\ \end{matrix} } \right|$. We want to rearrange its columns to match the form $\left| {\begin{matrix} 1 & x & {{x^2}} \\ 1 & y & {{y^2}} \\ 1 & z & {{z^2}} \\ \end{matrix} } \right|$.

• Why this step? By making both determinants identical, we can factor out a common determinant term, simplifying the equation significantly.

1. Swap Column 1 ($C_1$) and Column 3 ($C_3$): This changes the sign of the determinant. $D_1 = - \left| {\begin{matrix} 1 & {{a^2}} & a \\ 1 & {{b^2}} & b \\ 1 & {{c^2}} & c \\ \end{matrix} } \right|$
2. Swap Column 2 ($C_2$) and Column 3 ($C_3$): This again changes the sign. Since we performed two swaps, the net change in sign is $(-1) \times (-1) = 1$. $D_1 = (-1)^2 \left| {\begin{matrix} 1 & a & {{a^2}} \\ 1 & b & {{b^2}} \\ 1 & c & {{c^2}} \\ \end{matrix} } \right| = \left| {\begin{matrix} 1 & a & {{a^2}} \\ 1 & b & {{b^2}} \\ 1 & c & {{c^2}} \\ \end{matrix} } \right|$ So, the first determinant is equivalent to the standard Vandermonde determinant.

Step 4: Substitute the transformed determinant back and factor. Substitute the transformed $D_1$ back into equation (2): $\left| {\begin{matrix} 1 & a & {{a^2}} \\ 1 & b & {{b^2}} \\ 1 & c & {{c^2}} \\ \end{matrix} } \right| + abc \left| {\begin{matrix} 1 & a & {{a^2}} \\ 1 & b & {{b^2}} \\ 1 & c & {{c^2}} \\ \end{matrix} } \right| = 0$ Now, we can factor out the common determinant: $\left( 1 + abc \right) \left| {\begin{matrix} 1 & a & {{a^2}} \\ 1 & b & {{b^2}} \\ 1 & c & {{c^2}} \\ \end{matrix} } \right| = 0 \quad \text{... (3)}$

Step 5: Use the non-coplanar vector condition. We are given that the vectors $\left( {1,a,{a^2}} \right)$, $\left( {1,b,{b^2}} \right)$, and $\left( {1,c,{c^2}} \right)$ are non-coplanar.

• Why this step? The condition of non-coplanar vectors directly translates to the scalar triple product of these vectors being non-zero. The scalar triple product is precisely the determinant we have in equation (3). The scalar triple product of these vectors is: $\left| {\begin{matrix} 1 & a & {{a^2}} \\ 1 & b & {{b^2}} \\ 1 & c & {{c^2}} \\ \end{matrix} } \right|$ Since the vectors are non-coplanar, this determinant is not equal to zero: $\left| {\begin{matrix} 1 & a & {{a^2}} \\ 1 & b & {{b^2}} \\ 1 & c & {{c^2}} \\ \end{matrix} } \right| \neq 0$

Step 6: Solve for $abc$. From equation (3), we have the product of two terms equal to zero: $\left( 1 + abc \right) \left| {\begin{matrix} 1 & a & {{a^2}} \\ 1 & b & {{b^2}} \\ 1 & c & {{c^2}} \\ \end{matrix} } \right| = 0$ Since the second factor (the determinant) is non-zero (from Step 5), the first factor must be zero: $1 + abc = 0$ Solving for $abc$: $abc = -1$

Common Mistakes & Tips

• Sign Errors with Swaps: Be extremely careful when performing row or column swaps. Each swap introduces a factor of $-1$. An even number of swaps restores the original sign.
• Misinterpreting Non-Coplanar: The condition of non-coplanar vectors is critical. It directly implies that the determinant formed by the vectors' components is non-zero.
• Recognizing Vandermonde: Identifying the Vandermonde determinant structure early can simplify the problem significantly.

Summary

We utilized the property of splitting determinants and factored out common terms to simplify the given equation. By strategically swapping columns, we transformed one of the determinants into the standard Vandermonde form. The condition of non-coplanar vectors was used to establish that this Vandermonde determinant is non-zero. This led to the conclusion that the other factor in the simplified equation, $(1+abc)$, must be zero, yielding the product $abc = -1$.

The final answer is $\boxed{-1}$.