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Question: If a, b, c are all different and \(\left| \begin{matrix} a & a^{3} & a^{4} - 1 \\ b & b^{3} & b^{4} ...

If a, b, c are all different and ∣aa3a4−1bb3b4−1cc3c4−1∣\left| \begin{matrix} a & a^{3} & a^{4} - 1 \\ b & b^{3} & b^{4} - 1 \\ c & c^{3} & c^{4} - 1 \end{matrix} \right| = 0, then the value of abc(ab+bc+ca)abc(ab + bc + ca) is.

A

a+b+ca + b + c

B

0

C

a2+b2+c2a^{2} + b^{2} + c^{2}

D

a2−b2+c2a^{2} - b^{2} + c^{2}

Answer

a+b+ca + b + c

Explanation

Solution

[aa3a4−1bb3b4−1cc3c4−1]=0\begin{bmatrix} a & a^{3} & a^{4} - 1 \\ b & b^{3} & b^{4} - 1 \\ c & c^{3} & c^{4} - 1 \end{bmatrix} = 0or ∣aa3a4bb3b4cc3c4∣+∣aa3−1bb3−1cc3−1∣=0\left| \begin{matrix} a & a^{3} & a^{4} \\ b & b^{3} & b^{4} \\ c & c^{3} & c^{4} \end{matrix} \right| + \left| \begin{matrix} a & a^{3} & - 1 \\ b & b^{3} & - 1 \\ c & c^{3} & - 1 \end{matrix} \right| = 0

or abc ∣1a2a31b2b31c2c3∣+∣aa3−1a−ba3−b30a−ca3−c30∣=0abc\text{ }\left| \begin{matrix} 1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3} \end{matrix} \right| + \left| \begin{matrix} a & a^{3} & - 1 \\ a - b & a^{3} - b^{3} & 0 \\ a - c & a^{3} - c^{3} & 0 \end{matrix} \right| = 0

or abc ∣1a2a30a2−b2a3−b30a2−c2a3−c3∣+∣aa3−1a−ba3−b30(a−c)(a3−c3)0∣=0abc\text{ }\left| \begin{matrix} 1 & a^{2} & a^{3} \\ 0 & a^{2} - b^{2} & a^{3} - b^{3} \\ 0 & a^{2} - c^{2} & a^{3} - c^{3} \end{matrix} \right| + \left| \begin{matrix} a & a^{3} & - 1 \\ a - b & a^{3} - b^{3} & 0 \\ (a - c) & (a^{3} - c^{3}) & 0 \end{matrix} \right| = 0

or (abc)(a−b)(a−c)∣1a2a30a+ba2+b2+ab0a+ca2+c2+ac∣+(abc)(a - b)(a - c)\left| \begin{matrix} 1 & a^{2} & a^{3} \\ 0 & a + b & a^{2} + b^{2} + ab \\ 0 & a + c & a^{2} + c^{2} + ac \end{matrix} \right| +

a & a^{3} & - 1 \\ 1 & a^{2} + b^{2} + ab & 0 \\ 1 & a^{2} + c^{2} + ac & 0 \end{matrix} \right|$$ or $(a - b)(a - c)\lbrack(abc)\lbrack(a + b)(a^{2} + c^{2} + ac) -$ $$(a + c)(a^{2} + b^{2} + ab)\rbrack\rbrack + ( - 1)(a - b)(a - c)$$ $$\lbrack a^{2} + c^{2} + ac - a^{2} - b^{2} - ab\rbrack = 0$$ = $(abc)\lbrack(a - b)(a - c)(c - b)(ac + ab + bc)\rbrack$ $$+ ( - 1)(a - b)(a - c)(c - b)(a + b + c) = 0$$ $\Rightarrow$ $(abc)(ac + ab + bc) = a + b + c$.