Expand each of the following,using suitable identities: $(-2x + 3y + 2z)^2$

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(N/A) To expand $(-2x + 3y + 2z)^2$,we use the algebraic identity: $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.
Here,$a = -2x$,$b = 3y$,and $c = 2z$.
Substituting these values into the identity:
$(-2x + 3y + 2z)^2 = (-2x)^2 + (3y)^2 + (2z)^2 + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)$
Calculating each term:
$(-2x)^2 = 4x^2$
$(3y)^2 = 9y^2$
$(2z)^2 = 4z^2$
$2(-2x)(3y) = -12xy$
$2(3y)(2z) = 12yz$
$2(2z)(-2x) = -8zx$
Combining these results,we get:
$4x^2 + 9y^2 + 4z^2 - 12xy + 12yz - 8zx$

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