How do you factor completely $a^{2} - 121 b^{2}$ $a^2 - 121b^2$ ?

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Answer 1 · 2015-11-14T04:13:16

$(a + 11 b) (a - 11 b)$ $color(blue)((a+11b)(a-11b)$

$a^{2} - 121 b^{2}$ $a^2−121b^2$
This expression can be rewritten as ${(a)}^{2} - {(11 b)}^{2}$ $(a)^2−(11b)^2$

Now, as per the below property,
$x^{2} - y^{2} = (x + y) (x - y)$ $color(blue)(x^2-y^2=(x+y)(x-y)$ , we can rewrite the expression as:

${(a)}^{2} - {(11 b)}^{2} = (a + 11 b) (a - 11 b)$ $(a)^2−(11b)^2=color(blue)((a+11b)(a-11b)$
(Here, $x = a, y = 11 b$ $x=a,y=11b$ )

So the factorised form of the expression is:
$= (a + 11 b) (a - 11 b)$ $=color(blue)((a+11b)(a-11b)$

How do you factor completely a2−121b2a^2 - 121b^2?