if x³ + 3x² - 9x  + c is of the form (x-a)²(x-b) then find c

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$$\begin{gathered} {\mathrm{x}}^3+3{\mathrm{x}}^2+9\mathrm{x}+\mathrm{c}=0 \\ \mathrm{a} \mathrm{and} \mathrm{c} \mathrm{are} \mathrm{roots} \mathrm{of} \mathrm{the} \mathrm{equation}. \\ \mathrm{Sum} \mathrm{of} \mathrm{roots}=\mathrm{a}+\mathrm{a}+\mathrm{b}=-\frac{3}{1} \\ \Rightarrow 2\mathrm{a}+\mathrm{b}=-3 \\ \Rightarrow \mathrm{b}=-3-2\mathrm{a}=-\left(3+2\mathrm{a}\right) \\ \mathrm{Sum} \mathrm{of} \mathrm{roots} \mathrm{taking} \mathrm{two} \mathrm{at} \mathrm{a} \mathrm{time} \mathrm{we} \mathrm{get} \\ {\mathrm{a}}^2+\mathrm{ab}+\mathrm{ab}=\frac{9}{1} \\ \Rightarrow {\mathrm{a}}^2+2\mathrm{ab}=-9 \\ \Rightarrow {\mathrm{a}}^2-2\mathrm{a}\left(3+2\mathrm{a}\right)=9 \\ \Rightarrow {\mathrm{a}}^2-6\mathrm{a}-4{\mathrm{a}}^2=9 \\ \Rightarrow 3{\mathrm{a}}^2+6\mathrm{a}+9=0 \\ \mathrm{Divide} \mathrm{each} \mathrm{term} \mathrm{by} 3 \\ \Rightarrow {\mathrm{a}}^2+2\mathrm{a}+1=0 \\ \Rightarrow {\left(\mathrm{a}+1\right)}^2=0 \\ \Rightarrow \mathrm{a}=-1 \\ \mathrm{b}=-\left(3+2\mathrm{a}\right)=-\left(3-2\right)=-1 \\ \mathrm{Product} \mathrm{of} \mathrm{roots}=-\frac{\mathrm{c}}{1} \\ \left(-1\right)\times \left(-1\right)\times \left(-1\right)=-\mathrm{c} \\ \Rightarrow \mathrm{c}=1 \end{gathered}$$

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