Evaluate left|begin{array}{ccc}x & y & x+y y | vedclass

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(A) Let $\Delta = \left|\begin{array}{ccc}x & y & x+y \ y & x+y & x \ x+y & x & y\end{array}ight|$.Applying the row operation $R_{1} ightarrow R

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$-2(x^{3}+y^{3})$

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(A) Let $\Delta = \left|\begin{array}{ccc}x & y & x+y \ y & x+y & x \ x+y & x & y\end{array}ight|$.Applying the row operation $R_{1} ightarrow R_{1} + R_{2} + R_{3}$,we get:$\Delta = \left|\begin{array}{ccc}2(x+y) & 2(x+y) & 2(x+y) \ y & x+y & x \ x+y & x & y\end{array}ight|$.Taking $2(x+y)$ common from $R_{1}$:$\Delta = 2(x+y) \left|\begin{array}{ccc}1 & 1 & 1 \ y & x+y & x \ x+y & x & y\end{array}ight|$.Applying column operations $C_{2} ightarrow C_{2} - C_{1}$ and $C_{3} ightarrow C_{3} - C_{1}$:$\Delta = 2(x+y) \left|\begin{array}{ccc}1 & 0 & 0 \ y & x & x-y \ x+y & -y & -x\end{array}ight|$.Expanding along $R_{1}$:$\Delta = 2(x+y) [1 \cdot (x(-x) - (-y)(x-y))]$.$\Delta = 2(x+y) [-x^{2} + y(x-y)]$.$\Delta = 2(x+y) [-x^{2} + xy - y^{2}]$.$\Delta = -2(x+y)(x^{2} - xy + y^{2})$.Using the identity $a^{3} + b^{3} = (a+b)(a^{2} - ab + b^{2})$:$\Delta = -2(x^{3} + y^{3})$.

Evaluate $\left|\begin{array}{ccc}x & y & x+y \\ y & x+y & x \\ x+y & x & y\end{array}\right|$

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