$a^{36} - b^{36}$ is exactly divisible by

a - b

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a + b

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a^{2} + b^{2}

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All three

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Solution

The correct option is A $a - b$
Since, $a^{36} - b^{36}$
$= (a^{18})^{2} - (b^{18})^{2}$ [ $∵ a^{2} - b^{2} = (a - b) (a + b)$ ]
$= (a^{18} - b^{18}) (a^{18} + b^{18})$
$= [(a^{9})^{2} - (b^{9})^{2}] (a^{18} + b^{18})$
$= (a^{9} - b^{9}) (a^{9} + b^{9}) (a^{18} + b^{18})$
$= [(a^{3})^{3} - (b^{3})^{3}] (a^{9} + b^{9}) (a^{18} + b^{18})$ [ $∵ a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + a b)$ ]
$= (a^{3} - b^{3}) (a^{6} + b^{6} + a^{3} b^{3}) (a^{9} + b^{9}) (a^{18} + b^{18})$
$= (a - b) (a^{2} + b^{2} + a b) (a^{6} + b^{6} + a^{3} b^{3}) (a^{9} + b^{9}) (a^{18} + b^{18})$
Option A is correct.

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Similar questions

a^{36} + a^{36}

$P (n) : a^{2 n} - b^{2 n} is divisible by a + b, \forall n ϵ N$

To prove P(n) using mathematical induction, the base case is

a, b, c

a^{2} = b^{2} + c^{2} - 2 b c \sqrt{1 - a^{2}}, b^{2} = c^{2} + a^{2} - 2 c a \sqrt{1 - b^{2}}, c^{2} = a^{2} + b^{2} - 2 a b \sqrt{1 - c^{2}}

a = c \sqrt{1 - b^{2}} + b \sqrt{1 - c^{2}}

\frac{\sqrt{a^{2} - b^{2}} + a}{\sqrt{a^{2} + b^{2}} + b} \div \frac{\sqrt{a^{2} + b^{2}} - b}{a - \sqrt{a^{2} - b^{2}}}

a

b

a^{2} + b^{2}

10

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