Answer
$16(x^{4}+2x^{3}-x^{2}-2x+1)$
Work Step by Step
$(4x^{2}+4x-4)^{2} $
$=[4(x^{2}+x-1)]^{2} $
$=4^{2}(x^{2}+x-1)^{2} $
$=16(x^{2}+x-1)^{2} $
$=16(x^{2}+(x-1))^{2} $
Using the formula $(a+b)^{2}= a^{2}+2ab+b^{2}$
Let $a$ be $x^{2}$ and $b$ be $(x-1)$
$(x^{2})^{2}+2(x^{2})(x-1)+(x-1)^{2}$
$=16[x^{4}+2x^{2}(x-1)+(x-1)^{2}]$
[Using $(a-b)^{2}= a^{2}-2ab+b^{2}$
$(x-1)^{2}= x^{2}-2x+1 $]
$=16[x^{4}+2x^{3}-2x^{2}+x^{2}-2x+1]$
Combine like terms.
$= 16[x^{4}+2x^{3}-x^{2}-2x+1]$
