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Answer:$(5 x+\lambda)(\lambda-x)^{2}$

Hint: Use determinant formula

Given: $\left|\begin{array}{ccc} x+\lambda & 2 x & 2 x \\ 2 x & x+\lambda & 2 x \\ 2 x & 2 x & x+\lambda \end{array}\right|=(5 x+\lambda)(\lambda-x)^{2}$

Solution:

\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} x+\lambda & 2 x & 2 x \\ 2 x & x+\lambda & 2 x \\ 2 x & 2 x & x+\lambda \end{array}\right| \\ &\text { Use } \mathrm{C}_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &=\left|\begin{array}{ccc} 5 x+\lambda & 2 x & 2 x \\ 5 x+\lambda & x+\lambda & 2 x \\ 5 x+\lambda & 2 x & x+\lambda \end{array}\right| \end{aligned}

\begin{aligned} &(5 x+\lambda) \text { common from } \mathrm{C}_{1}\\ &=(5 x+\lambda)\left|\begin{array}{ccc} 1 & 2 x & 2 x \\ 1 & x+\lambda & 2 x \\ 1 & 2 x & x+\lambda \end{array}\right|\\ &\text { Now } \mathrm{R}_{2} \rightarrow R_{2}-R_{1} \text { and } \mathrm{R}_{3} \rightarrow R_{3}-R_{1}\\ &=(5 x+\lambda)\left|\begin{array}{ccc} 1 & 2 x & 2 x \\ 0 & x-\lambda & 0 \\ 0 & 0 & x-\lambda \end{array}\right|\\ &=(5 x+\lambda)(-1)(-1)\left|\begin{array}{ccc} 1 & 2 x & 2 x \\ 0 & -x+\lambda & 0 \\ 0 & 0 & -x+\lambda \end{array}\right| \end{aligned}

$\text { Expanding w.r.t } \mathrm{R}_{1}$

\begin{aligned} &=(5 x+\lambda)\left[1(\lambda-x)^{2}-0+0\right] \\ &=(5 x+\lambda)(\lambda-x)^{2} \\ &=R \cdot H . S \end{aligned}

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