Factorise
$a^{3} - a^{2} - a b + a + b - 1$

(a + 1) (a^{2} - b + 1)

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(a - 1) (a^{2} - b - 1)

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(a - 1) (a^{2} - b + 1)

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(a) (a^{2} - b + 1)

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Solution

The correct option is B $(a - 1) (a^{2} - b + 1)$
$F a c t o r i z i n g a^{3} - a^{2} - a b + a + b - 1, r e a r r a n g e a n d, = a^{2} (a - 1) - b (a - 1) + 1 (a - 1) = (a - 1) (a^{2} - b + 1)$

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Similar questions

Q. If

a^{2} - 4 a + 1 = 4

, then the value of

\frac{a^{3} - a^{2} + a - 1}{a^{2} - 1} (a^{2} \neq 1)

is equal to

Q. Add

(a + 1) (a^{2} - a + 1) a n d (a - 1) (a^{2} + a + 1);

Q. If p be the length of the perpendicular from the origin on the line x/a + y/b = 1, then
(a) p² = a² + b²

(b)

p^{2} = \frac{1}{a^{2}} + \frac{1}{b^{2}}

(c)

\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}

(d) none of these

Factorise : $a^{3} - a^{2} - a b + a + b - 1$

If $Δ = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣$ $A_{i}, B_{i}, C_{i}$ are respectively are cofator of $a_{i}, b_{i}, c_{i}$ -- for all i = 1, 2, 3, then

$∣ ∣ ∣ ∣ \begin{matrix} A_{1} & B_{1} & C_{1} A_{2} & B_{2} & C_{2} A_{3} & B_{3} & C_{3} \end{matrix} ∣ ∣ ∣ ∣$

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Factorisea3−a2−ab+a+b−1

The correct option is B (a−1)(a2−b+1)Factorizinga3−a2−ab+a+b−1,rearrangeand,=a2(a−1)−b(a−1)+1(a−1)=(a−1)(a2−b+1)

Factorise
$a^{3} - a^{2} - a b + a + b - 1$

The correct option is B $(a - 1) (a^{2} - b + 1)$
$F a c t o r i z i n g a^{3} - a^{2} - a b + a + b - 1, r e a r r a n g e a n d, = a^{2} (a - 1) - b (a - 1) + 1 (a - 1) = (a - 1) (a^{2} - b + 1)$