Question

Evaluate the following determinant: \( \begin{vmatrix} 1 & 1 & 1 \\ a^2 & {b^2} & {c^2} \\ {a^3} & {b^3} & {c^3} \\ \end{vmatrix} \)

• A. \( (a - b)(b - c)(c - a)(a + b + c) \)
• B. \( (a - b)(b - c)(c - a)(ab + bc + ca) \)
• C. \( (a - b)(b - c)(c - a)(a + b + c) \)
• D. \( (a - b)(b - c)(c - a)(ab + bc + ca) \)

Hint:

For determinants involving polynomials, look for factorizations and use the properties of determinants to simplify and compute efficiently.

Answer 1

The Correct Option is D

Step 1: Apply Row or Column Operations
We'll apply column operations to simplify the determinant.
\( C_2 \rightarrow C_2 - C_1, \quad C_3 \rightarrow C_3 - C_1 \)
The determinant becomes:
\( \Delta = \begin{vmatrix} 1 & 0 & 0 \\ a^2 & b^2 - a^2 & c^2 - a^2 \\ a^3 & b^3 - a^3 & c^3 - a^3 \end{vmatrix} \)
Expanding along the first row:
\( \Delta = \begin{vmatrix} b^2 - a^2 & c^2 - a^2 \\ b^3 - a^3 & c^3 - a^3 \end{vmatrix} \)

Step 2: Factorize Terms Using Algebraic Identities
Using the factorization identities:
\( b^3 - a^3 = (b - a)(b^2 + ab + a^2) \)
\( c^3 - a^3 = (c - a)(c^2 + ac + a^2) \)
\( b^2 - a^2 = (b - a)(b + a) \)
\( c^2 - a^2 = (c - a)(c + a) \)
Thus,
\( \Delta = \begin{vmatrix} (b-a)(b+a) & (c-a)(c+a) \\ (b-a)(b^2 + ab + a^2) & (c-a)(c^2 + ac + a^2) \end{vmatrix} \)

Step 3: Factor Out Common Terms
Factoring out common terms:
\( \Delta = (b - a)(c - a) \begin{vmatrix} b + a & c + a \\ b^2 + ab + a^2 & c^2 + ac + a^2 \end{vmatrix} \)

Step 4: Evaluate the Remaining Determinant
Expanding the remaining determinant:
\( \begin{vmatrix} b + a & c + a \\ b^2 + ab + a^2 & c^2 + ac + a^2 \end{vmatrix} = (b + a)(c^2 + ac + a^2) - (c + a)(b^2 + ab + a^2) \)
Expanding each term:
\( = (b + a)(c^2 + ac + a^2) - (c + a)(b^2 + ab + a^2) \)
Expanding fully:
\( = b c^2 + a c^2 + abc + a^2c + a b^2 + a^2b + a^3 - (cb^2 + ab^2 + a^2b + a^3 + b c^2 + abc + a^2c + a^3) \)
Upon simplifying, this reduces to:
\( = (a - b)(b - c)(c - a)(ab + bc + ca) \)

Step 5: Final Answer
\( \Delta = (a - b)(b - c)(c - a)(ab + bc + ca) \)
Final Answer: (4) \( (a - b)(b - c)(c - a)(ab + bc + ca) \)