$a^{3}$ - $b^{3}$ is always divisible by:

a - b

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a × b

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a + b

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2a + b

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Solution

The correct option is A
a - b

Take $a = 2$ and $b = 1$ and check.
Alternatively, we know the algebraic identity,

$a^{3} - b^{3} = (a - b) (a^{2} + 2 a b + b^{2}$ )

Hence , we can see that $(a - b)$ is a factor of $a^{3}$ - $b^{3}$
Thus, $a^{3}$ - $b^{3}$ is divisible by $(a - b)$

Suggest Corrections

Similar questions

a^{8} - 2 a^{4} b^{4} + b^{8}

In a much-simplified form, the given algebraic expression could be written as: $(a^{2} + a b + b^{2}) (a - b)$

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

a^{3} + b^{3} + c^{3} - a^{2} b + a b^{2} - b^{2} c + c^{2} b + a^{2} c - a c^{2}

A

B

A^{3} = B^{3}

A^{2} B = B^{2} A

d e t (A^{2} + B^{2})

(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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