Expand $(4a - 2b - 3c)^2.$
(N/A) Using the algebraic identity $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$,we have:
$(4a - 2b - 3c)^2 = [4a + (-2b) + (-3c)]^2$
$= (4a)^2 + (-2b)^2 + (-3c)^2 + 2(4a)(-2b) + 2(-2b)(-3c) + 2(-3c)(4a)$
$= 16a^2 + 4b^2 + 9c^2 - 16ab + 12bc - 24ac$
$(4a - 2b - 3c)^2 = [4a + (-2b) + (-3c)]^2$
$= (4a)^2 + (-2b)^2 + (-3c)^2 + 2(4a)(-2b) + 2(-2b)(-3c) + 2(-3c)(4a)$
$= 16a^2 + 4b^2 + 9c^2 - 16ab + 12bc - 24ac$
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