Answer
$7$
Work Step by Step
Recall, in order to solve problems involving order of operations, we use the PEMDAS rule.
First Priority: P - parentheses and other grouping symbols (including fraction bars)
Second Priority: E - exponents
Third Priority: M/D - Multiplication or division, whichever comes first from the left to the right
Fourth Priority: A/S - Addition or subtraction, whichever comes first from the left to the right
We follow order of operations to obtain that the expression, $
\dfrac{3(4-2)^2}{4}+\dfrac{2(-1+3)^3}{4}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{3(4-2)(4-2)}{4}+\dfrac{2(-1+3)(-1+3)(-1+3)}{4}
\\\\=
\dfrac{3(2)(2)}{4}+\dfrac{2(2)(2)(2)}{4}
\\\\=
\dfrac{3(\cancel{2})(\cancel{2})}{\cancel{2}(\cancel{2})}+\dfrac{\cancel{2}(\cancel{2})(2)(2)}{\cancel{2}(\cancel{2})}
\\\\=
3+4
\\\\=
7
.\end{array}
