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How do you multiply $(6 - [3 \sqrt{2}]) (6 + [3 \sqrt{2}])$ $(6-[3sqrt2])(6+[3sqrt2])$ ?

1 Answer

Jul 24, 2015

$(6 - [3 \sqrt{2}) (6 + [3 \sqrt{2}])$ $(6-[3sqrt(2])(6+[3sqrt(2)])$
$XXXX$ $color(white)("XXXX")$ $XXXX$ $color(white)("XXXX")$ $XXXX$ $color(white)("XXXX")$ $XXXX$ $color(white)("XXXX")$ $= 18$ $= 18$

Explanation:

In general
$XXXX$ $color(white)("XXXX")$ $(a + b) (a - b) = a^{2} - b^{2}$ $(a+b)(a-b) = a^2-b^2$
(this is commonly referred to as the difference of quares)

So
$XXXX$ $color(white)("XXXX")$ $(6 - [3 \sqrt{2}) (6 + [3 \sqrt{2}])$ $(6-[3sqrt(2])(6+[3sqrt(2)])$

$XXXX$ $color(white)("XXXX")$ $XXXX$ $color(white)("XXXX")$ $= 6^{2} - {[3 \sqrt{2}]}^{2}$ $= 6^2 - [3sqrt(2)]^2$

$XXXX$ $color(white)("XXXX")$ $XXXX$ $color(white)("XXXX")$ $= 36 - 18$ $= 36 - 18$

$XXXX$ $color(white)("XXXX")$ $XXXX$ $color(white)("XXXX")$ $= 18$ $= 18$

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