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In simplifying $$[3a(b-c)+5][-3a(b-c)-5],$$ I used $$4(au+bv)(cu+dv)=acu^2+(ad+bc)uv+bdv^2.$$

I failed to apply the formula to the equation because $a=3a$, $b=-3a$, $c=5$, $d=-5$, $u=(b-c)$, $v=?$

There's no value of $v$ so I tried to find other special product but it's only

$$(au+bv)(cu+dv)=acu^2+(ad+bc)uv+bdv^2$$ fit to the equation.

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One factor's the negative of the other. You know how to square a binomial? – J. M. Jul 14 '12 at 7:42
up vote 1 down vote accepted

Hint: Let $x=3a(b-c)$ and $y=5$. Then \begin{align}[3a(b-c)+5][-3a(b-c)-5]&=(x+y)(-x-y)\\\\ &=(x+y)\cdot(-1)(x+y)\\\\ &=-(x+y)^2\\\\ &=-x^2-2xy-y^2. \end{align}

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