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(5c+2/3)(5c-2/3)
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((15c+2)(15c-2))/9
I don't get it lol
(5c*3+2)/3 * (5c*3-2)/3
so it will become (15c+2)/3*(15c-2)/3
=[(15c+2)(15c-2)]/9
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now u need to solve these two brackets by multiplyin them
15c^2+4?
no (15c)^2-4
\[(5c+\frac{2}{3})(5c-\frac{2}{3})\]
hmm.. \((a+b)(a-b\)) = a^2-b^2
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use (a+b)(a-b)
\[225c ^{2}-4\]/6
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