One of the factors of $a^{3} - b^{3} - a^{2} b + a b^{2} + a^{2} - b^{2}$ is

a + b

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b - a

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a - b

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a^{2} + b^{2}

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Solution

The correct option is C $a - b$
$a^{3} - b^{3} - a^{2} b + a b^{2} + a^{2} - b^{2} = (a^{3} - b^{3}) - a^{2} b + a b^{2} + (a^{2} - b^{2}) = [(a - b) (a^{2} + a b + b^{2})] + [(- a b) (a - b)] + [(a - b) (a + b)]$
$(∵ x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2}), x^{2} - y^{2} = (x + y) (x - y)) = (a - b) [(a^{2} + a b + b^{2}) - a b + (a + b)] = (a - b) [a^{2} + a b + b^{2} - a b + a + b] = (a - b) (a^{2} + b^{2} + a + b)$
We observe that $(a - b)$ is the common factor of the terms $(a^{3} - b^{3}), (- a^{2} b + a b^{2})$ and $(a^{2} - b^{2})$
Hence, one of the factors is $(a - b)$ .

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(a + b) + (a^{2} + a b + b^{2}) + (a^{3} + a^{2} b + a b^{2} + b^{3}) + . . . . .

a \neq 1, b \neq 1 a n d a \neq b

\frac{a^{2} - b^{2}}{a - b} - \frac{a^{3} - b^{3}}{a^{2} - b^{2}}

If a, b, c are in G.P., prove that:

(i) $a (b^{2} + c^{2}) = c (a^{2} + b^{2})$

(ii) $a^{2} b^{2} c^{2} (\frac{1}{a^{3}} + \frac{1}{b^{3}} + \frac{1}{c^{3}}) = a^{3} + b^{3} + c^{3}$

(iii) $\frac{(a + b + c)^{2}}{a^{2} + b^{2} + c^{2}} = \frac{a + b + c}{a - b + c}$

(iv) $\frac{1}{a^{2} - b^{2}} + \frac{1}{b^{2}} = \frac{1}{b^{2} - c^{2}}$

(v) $(a + 2 b + 2 c) (a - 2 b + 2 c) = a^{2} + 4 c^{2}$ .

a - a^{3} + b^{3} - b

(i) (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a)

(i i) (a + b)^{3} = a^{3} + b^{3} + 3 a b (a - b)

(i i i) (a - b)^{3} = a^{3} - b^{3} - 3 a b (a + b)

(i v) a^{2} - b^{2} = (a + b) (a - b)

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