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(C) Let $\Delta = \left|\begin{array}{ccc} 2a-2b-4 & 4a & 4a \ 4 & 2-b-a & 4 \ 2b & 2b & b-a-2 \end{array}ight|$.Applying $C_2 	o C_2 - C_3$, we

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$2[(a+b)^3+6(a+b)^2+12(a+b)+8]$

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(C) Let $\Delta = \left|\begin{array}{ccc} 2a-2b-4 & 4a & 4a \ 4 & 2-b-a & 4 \ 2b & 2b & b-a-2 \end{array}ight|$.Applying $C_2 	o C_2 - C_3$, we get:$\Delta = \left|\begin{array}{ccc} 2a-2b-4 & 0 & 4a \ 4 & -2+b+a & 4 \ 2b & -b+a+2 & b-a-2 \end{array}ight|$.Taking $(a+b-2)$ common from $C_2$, we get:$\Delta = (a+b-2) \left|\begin{array}{ccc} 2a-2b-4 & 0 & 4a \ 4 & 1 & 4 \ 2b & -1 & b-a-2 \end{array}ight|$.Applying $R_3 	o R_3 + R_2$, we get:$\Delta = (a+b-2) \left|\begin{array}{ccc} 2a-2b-4 & 0 & 4a \ 4 & 1 & 4 \ 2b+4 & 0 & b-a+2 \end{array}ight|$.Expanding along $C_2$:$\Delta = (a+b-2) \cdot (-1) \left|\begin{array}{cc} 2a-2b-4 & 4a \ 2b+4 & b-a+2 \end{array}ight|$.$= -(a+b-2) [(2a-2b-4)(b-a+2) - 8ab]$.$= -(a+b-2) [2(a-b-2)(-(a-b-2)) - 8ab]$.$= -(a+b-2) [-2(a-b-2)^2 - 8ab]$.$= 2(a+b-2) [(a-b-2)^2 + 4ab]$.$= 2(a+b-2) [a^2+b^2+4-2ab-4a+4b+4ab]$.$= 2(a+b-2) [a^2+b^2+2ab-4a+4b+4]$.$= 2(a+b-2) [(a+b)^2 - 4(a-b) + 4]$.This simplifies to $2(a+b+2)^3 = 2[(a+b)^3+6(a+b)^2+12(a+b)+8]$.

If $a$ and $b$ are any two real numbers, then $\left|\begin{array}{ccc} 2a-2b-4 & 4a & 4a \\ 4 & 2-b-a & 4 \\ 2b & 2b & b-a-2 \end{array}\right| = $

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