Question
From,
$p^3-4+3 p^2$, take away $5 p^2-3 p^3+p-6$
$p^3-4+3 p^2$, take away $5 p^2-3 p^3+p-6$
✓
Answer
Required expression:
$=\left(p^3-4+3 p^2\right)-\left(5 p^2-3 p^3+p-6\right)$
$=p^3-4+3 p^2-5 p^2+3 p^3-p+6$
$=p^3+3 p^3+3 p^2-5 p^2-p-4+6$
$=4 p^3-2 p^2-p+2$
$=\left(p^3-4+3 p^2\right)-\left(5 p^2-3 p^3+p-6\right)$
$=p^3-4+3 p^2-5 p^2+3 p^3-p+6$
$=p^3+3 p^3+3 p^2-5 p^2-p-4+6$
$=4 p^3-2 p^2-p+2$
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