Answer
$27x^{5}-9x^{4}+3x^{2}-3=$
$=(3x^{2}-3x+1)(9x^{3}+6x^{2}+3x+2)+(3x-5)$
Work Step by Step
$\left[\begin{array}{l}
\\\\
3x^{2}-3x+1\ )\\
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\\
\end{array}\right. \left.\begin{array}{llllll}
9x^{3} & +6x^{2} & +3x & +2 & & \\
\hline 27x^{5} & -9x^{4} & & +3x^{2} & & -3\\
27x^{5} & -27x^{4} & 9x^{3} & & & \\
-- & -- & -- & & & \\
& 18x^{4} & -9x^{3} & +3x^{2} & & -3\\
& 18x^{4} & -18x^{3} & +6x^{2} & & \\
& -- & -- & -- & & \\
& & 9x^{3} & -3x^{2} & & -3\\
& & 9x^{3} & -9x^{2} & 3x & \\
& & -- & -- & -- & \\
& & & 6x^{2} & -3x & -3\\
& & & 6x^{2} & -6x & +2\\
& & & -- & -- & --\\
& & & & 3x & -5
\end{array}\right]$
$Q(x)=9x^{3}+6x^{2}+3x+2,\quad R(x)=3x-5$
In the form $P(x)=D(x)\cdot Q(x)+R(x),$
$27x^{5}-9x^{4}+3x^{2}-3=$
$=(3x^{2}-3x+1)(9x^{3}+6x^{2}+3x+2)+(3x-5)$
