Expand $(4a -2b -3c)^2.$
Expand $(4a -2b -3c)^2.$
Using Identity $V$, we have
$(4 a-2 b-3 c)^{2}=[4 a+(-2 b)+(-3 c)]^{2}$
$=(4 a)^{2}+(-2 b)^{2}+(-3 c)^{2}+2(4 a)(-2 b)+2(-2 b)(-3 c)+2(-3 c)(4 a)$
$=16 a^{2}+4 b^{2}+9 c^{2}-16 a b+12 b c-24 a c$
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