Factorise:
$a b^{2} + (a - 1) b - 1$

(b + 1) (a - 1)

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(b + 1) (b - 1)

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(b + 1) (a b - 1)

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(b - 1) (a b - 1)

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Solution

The correct option is C $(b + 1) (a b - 1)$
$a b^{2} + (a - 1) b - 1$
$= a b^{2} + a b - b - 1$
$= a b (b + 1) - 1 (b + 1)$
$= (b + 1) (a b - 1)$

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Similar questions

\frac{a + b + c}{a^{- 1} b^{- 1} + b^{- 1} c^{- 1} + c^{- 1} a^{- 1}} = a b c

{(a^{- 1} + b^{- 1})}^{- 1} = \frac{a b}{a + b}

\frac{a}{(a x - 1)} + \frac{b}{(b x - 1)} = (a + b), (x \neq \frac{1}{a}, \frac{1}{b})

[A (A + B)^{- 1} B]^{- 1} = B^{- 1} + A^{- 1}

[A (A + B)^{- 1} B]^{- 1} = [A (A^{- 1} + B^{- 1}) B]^{- 1} = [(I + A B^{- 1}) B]^{- 1} = [(B + A B^{- 1} B)]^{- 1} = [(B + A I)]^{- 1} = [(B + 1)]^{- 1} = B^{- 1} + A^{- 1}

Prove that:

(i) $\frac{a + b + c}{a^{- 1} b^{- 1} + b^{- 1} c^{- 1} + c^{- 1} a^{- 1}} = a b c$

(ii) $(a^{- 1} + b^{- 1})^{- 1} = \frac{a b}{a + b}$

(a + c) \neq 0

∣ ∣ ∣ ∣ \begin{matrix} a & a + 1 & a - 1 - b & b + 1 & b - 1 c & c - 1 & c + 1 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ ∣ \begin{matrix} a + 1 & b + 1 & c - 1 a - 1 & b - 1 & c + 1 (- 1)^{n + 2} a & (- 1)^{n + 1} b & (- 1)^{n} c \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

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