How do you multiply $(\sqrt{6} + 6 c) (\sqrt{6} - 6 c)$ $(sqrt6 +6c)(sqrt6 -6c)$ ?

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Answer 1 · 2015-06-05T20:17:35

You can recognise this as being of the form $(a + b) (a - b) = a^{2} - b^{2}$ $(a+b)(a-b) = a^2-b^2$

So:

$(\sqrt{6} + 6 c) (\sqrt{6} - 6 c) = {(\sqrt{6})}^{2} - {(6 c)}^{2} = 6 - 36 c^{2}$ $(sqrt(6)+6c)(sqrt(6)-6c) = (sqrt(6))^2-(6c)^2=6-36c^2$

How do you multiply (sqrt6 +6c)(sqrt6 -6c) (√6+6c)(√6−6c) (sqrt6 +6c)(sqrt6 -6c) ?