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Complete the following without the use of a calculator:

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$\\var{n[0]}^2$ = [[0]]

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Here the square is acting only on the number and not on the negative sign. So you square the number first and then take the negative. That is, $\\var{n[0]}^2=-(\\var{-n[0]})^2=-(\\var{-n[0]}\\times\\var{-n[0]})=-\\var{-ans1}$.

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Note

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\\[\\var{n[0]}^2\\ne (\\var{n[0]})^2\\]

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$(\\var{n[1]})^2$ = [[0]]

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Here we have the square acting on a negative number. When two negatives are multiplied together the result is a positive number. 

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This means when a negative number is squared the result is positive.

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In particular, $(\\var{n[1]})^2=(\\var{n[1]})\\times(\\var{n[1]})=\\var{ans2}$.

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Note

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\\[\\var{n[1]}^2\\ne (\\var{n[1]})^2\\]

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$-(\\var{n[1]})^2$ = [[0]]

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Deal with the squaring of the negative number first (which results in a positive) and then deal with the negative outside the brackets. 

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That is, $-(\\var{n[1]})^2=-(\\var{n[1]}\\times\\var{n[1]})=-(\\var{ans2})=\\var{-ans2}$

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Note

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\\[-(\\var{n[1]})^2\\ne \\var{-n[1]}^2\\]

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$(\\var{m})^3$ = [[0]]

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We have $(\\var{m})^3$. Recall this is the same as $(\\var{m})\\times(\\var{m})\\times(\\var{m})$. Here we have three negative symbols, two give a positive and the remaining negative makes the result negative, that is,

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\\[(\\var{m})^3=\\var{ans3}\\]

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$(-1)^\\var{p}$ = [[0]]

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Any even power of a negative number will have an even number of negatives multiplied together. This will then result in a positive number.

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Any odd power of a negative number will have an odd number of negatives multiplied together. This will then result in a negative number.

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