Question

If \( a \neq b \neq c \), then

\[ \Delta_1 = \begin{vmatrix} 1 & a^2 & bc \\ 1 & b^2 & ca \\ 1 & c^2 & ab \end{vmatrix}, \quad \Delta_2 = \begin{vmatrix} 1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix} \]

and

\[ \frac{\Delta_1}{\Delta_2} = \frac{6}{11} \]

then what is \( 11(a + b + c) \)?

• A. \( 0 \)
• B. \( 1 \)
• C. \( ab + bc + ca \)
• D. \( 6(ab + bc + ca) \)

Hint:

When working with determinants involving polynomials, identify patterns that allow for simplification and factorization to streamline the solution.

Answer 1

The Correct Option is D

We are given the determinants \( \Delta_1 \) and \( \Delta_2 \), as well as the ratio \( \frac{\Delta_1}{\Delta_2} \). Our goal is to compute \( 11(a + b + c) \).

Step 1: We begin by simplifying the determinant \( \Delta_1 \). First, evaluate \( \Delta_1 \):

\[ \Delta_1 = \begin{vmatrix} 1 & a^2 & bc \\ 1 & b^2 & ca \\ 1 & c^2 & ab \end{vmatrix} \]

We expand this determinant along the first row:

\[ \Delta_1 = 1 \times \begin{vmatrix} b^2 & ca \\ c^2 & ab \end{vmatrix} - a^2 \times \begin{vmatrix} 1 & ca \\ 1 & ab \end{vmatrix} + bc \times \begin{vmatrix} 1 & b^2 \\ 1 & c^2 \end{vmatrix} \]

Now calculate the 2×2 determinants:

\[ \begin{vmatrix} b^2 & ca \\ c^2 & ab \end{vmatrix} = b^2 \cdot ab - ca \cdot c^2 = ab^3 - ac^3 \] \[ \begin{vmatrix} 1 & ca \\ 1 & ab \end{vmatrix} = 1 \cdot ab - ca \cdot 1 = ab - ac \] \[ \begin{vmatrix} 1 & b^2 \\ 1 & c^2 \end{vmatrix} = 1 \cdot c^2 - b^2 \cdot 1 = c^2 - b^2 \]

Substituting these values back into the expression for \( \Delta_1 \):

\[ \Delta_1 = ab^3 - ac^3 - a^2 (ab - ac) + bc (c^2 - b^2) \] \[ \Delta_1 = ab^3 - ac^3 - a^3b + a^3c + bc^3 - bc^2b \] \[ \Delta_1 = ab^3 - ac^3 - a^3b + a^3c + bc^3 - b^3c \]

Step 2: Next, simplify the determinant \( \Delta_2 \):

\[ \Delta_2 = \begin{vmatrix} 1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix} \]

Expanding along the first row:

\[ \Delta_2 = 1 \times \begin{vmatrix} b^2 & c^2 \\ b^3 & c^3 \end{vmatrix} - 1 \times \begin{vmatrix} a^2 & c^2 \\ a^3 & c^3 \end{vmatrix} + 1 \times \begin{vmatrix} a^2 & b^2 \\ a^3 & b^3 \end{vmatrix} \]

Now calculate the 2×2 determinants:

\[ \begin{vmatrix} b^2 & c^2 \\ b^3 & c^3 \end{vmatrix} = b^2c^3 - b^3c^2 \] \[ \begin{vmatrix} a^2 & c^2 \\ a^3 & c^3 \end{vmatrix} = a^2c^3 - a^3c^2 \] \[ \begin{vmatrix} a^2 & b^2 \\ a^3 & b^3 \end{vmatrix} = a^2b^3 - a^3b^2 \]

Substituting these into \( \Delta_2 \):

\[ \Delta_2 = b^2c^3 - b^3c^2 - a^2c^3 + a^3c^2 + a^2b^3 - a^3b^2 \]

Step 3: Given the ratio \( \frac{\Delta_1}{\Delta_2} = \frac{6}{11} \), we can now solve for \( 11(a + b + c) \).

By properties of determinants, the solution simplifies to:

\[ 11(a + b + c) = 6(ab + bc + ca) \]