Final answer to the problem
$f\left(x\right)=\frac{x^4\left(x^2-4x+4\right)}{\left(x^2+7\right)^5}$
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Step-by-step Solution
1
A binomial squared (difference) is equal to the square of the first term, minus the double product of the first by the second, plus the square of the second term. In other words: $(a-b)^2=a^2-2ab+b^2$
$f\left(x\right)=\frac{x^4\left(x^2-4x+4\right)}{\left(x^2+7\right)^5}$
Final answer to the problem
$f\left(x\right)=\frac{x^4\left(x^2-4x+4\right)}{\left(x^2+7\right)^5}$
