bbc - b - cabb - a - bca

bcab

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acbb

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abcb

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cabb

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None of the above

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Solution

The correct option is C abcb
The pattern is bbca $∣$ bbca $∣$ bbca $∣$ bbca

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\frac{A B}{A D}

∣ ∣ ∣ ∣ \begin{matrix} b + c & a & b c + a & c & a a + b & b & c \end{matrix} ∣ ∣ ∣ ∣ = (a + b + c) (a - c)^{2}

a b

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