# [(4(3−1)^{3})−2^{3}]−[(5×2)^{2}]

$\left[\left(4{\left(3-1\right)}^{3}\right)-{2}^{3}\right]-\left[{\left(5×2\right)}^{2}\right]$
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The order of operations is PEMDAS which states that the operations must be evaluated in the order of Parentheses, Exponents, Multiplication and Division in order from left to right, and then Addition and Subtraction in order from left to right.
To simplify $\left[\left(4{\left(3-1\right)}^{3}\right)-{2}^{3}\right]-\left[{\left(5×2\right)}^{2}\right]$ we must then evaluate the expression $3-1$ and $5×2$ first since they are inside of parentheses.
Since $3-1=2$ and $5×2=10$, then:
$\left[\left(4{\left(3-1\right)}^{3}\right)-{2}^{3}\right]-\left[{\left(5×2\right)}^{2}\right]=\left[4{\left(2\right)}^{3}-\left({2}^{3}\right)\right]-\left[{\left(10\right)}^{2}\right]$
Next we need to evaluate the exponents. Since ${2}^{3}={8}^{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{10}^{2}=100$, then: $\left[4{\left(2\right)}^{3}-\left({2}^{3}\right)\right]-\left[{\left(10\right)}^{2}\right]=\left[4\left(8\right)-8\right]-100$
The next operation we must evaluate is then the multiplication. Since $4\left(8\right)=324\left(8\right)=32$, then:
$\left[4\left(8\right)-8\right]-100=\left(32-8\right)-100$
The remaining operation to evaluate is the subtraction. Subtract 32 and 8 and then subtract 100 from this result:
$\left(32-8\right)-100=24-100=-76$