Q44 / 160 · AP EAMCET 2024 (19 May Shift 2) · Mathematics

$\left|\begin{array}{ccc}1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3\end{array}\right|=$

  • A
    $(a-b)(b-c)(c-a)(a+b+c)$
  • B
    $(a-b)(b-c)(c-a)$
  • C
    $(a-b)(b-c)(a-c)(a b+b c+c a)$
  • Correct answer
    D
    $(a-b)(b-c)(c-a)(a b+b c+c a)$

Solution

Since, $\left|\begin{array}{ccc}1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3\end{array}\right|$ $=\left|\begin{array}{ccc}0 & 1 & 0 \\ a^2-b^2 & b^2 & c^2-b^2 \\ a^3-b^3 & b^3 & c^3-b^3\end{array}\right| \quad\binom{\because C_1 \rightarrow C_1-C_2}{C_3 \rightarrow C_3-C_2}$ $=(a-b)(c-b)\left|\begin{array}{ccc}0 & 1 & 0 \\ a+b & b^2 & b+c \\ a^2+b^2+a b & b^3 & b^2+c^2+b c\end{array}\right|$ $=(a-b)(c-b)\left|\begin{array}{ccc}0 & 1 & 0 \\ a+b & b^2 & c-a \\ a^2+b^2+a b & b^3 & c^2+b c-a^2-a b\end{array}\right|\left(\because C_3 \rightarrow C_3-C_1\right)$ $\begin{aligned} & =(a-b)(c-b)(c-a)\left|\begin{array}{ccc}0 & 1 & 0 \\ a+b & b^2 & 1 \\ a^2+b^2+a b & b^3 & a+b+c\end{array}\right| \\ & =-(a-b)(c-b)(c-a)\left\{(a+b)(a+b+c)-\left(a^2+b^2+a b\right)\right\} \\ & =(a-b)(b-c)(c-a)(a b+b c+a c) .\end{aligned}$

Hence, the correct answer is option: 'D'.

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