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The given equation is$y={(x-5)}^3-4$.

The aim of the problem is to find how many units to the left/right must we shift the graph of $y=x^3$and then how many units up/down must we shift the graph of$y=x^3.$

• The graph of $y=f(x-h)$, where $h>0$is same as the graph of localid="1647317716779" $f$shifted horizontally right localid="1647312983714" $h$units.
• The graph of localid="1647317634014" $y=f(x+h)$, where $h>0$is same as the graph of localid="1647317725686" $f$shifted horizontally left localid="1647313021400" $h$units.
• The graph of the function localid="1647313030841" $y=f\left(x\right)+k$, where $k>0$is same as the graph of localid="1647317731879" $f$shifted vertically uplocalid="1647317697204" $k$units.
• The graph of the function localid="1647313057733" $y=f\left(x\right)-k$, where $k<0$is same as the graph of localid="1647317677613" $f$shifted vertically down localid="1647317706576" $k$units.

Since the argument $x$of the function $y=x^3$is been replaced by $x-5$where $5>0$, the graph of the function ${(x-5)}^3$will be shifted horizontally right by $5$units.

Further, since $4$is subtracted from ${(x-5)}^3$, the graph of the functionlocalid="1647966795614" $y={(x-5)}^3$will be shifted vertically down $4$units.

Therefore, to grpah $y={(x-5)}^3-4,$shift the graph of $y=x^3$to the right$5$units and then down$4$units.