Using properties of determinants, show the following: (b+c)^2& ab & ca ab & (b+c)^2 & bc ac & bc & (a+b)^2 =2abc (a+b+c)^3

= (b+c)^2& ab & ca ab & (b+c)^2 & bc ac & bc & (a+b)^2 ; Applying R_1 R_1, R_2 R_2, R_3 R_3 we get=1 a b c a(b+c)^2& b a^2 & c a^2 a b^2 & b(c+a)^2 & c b^2 a c^2 & b c^2 &c (a+b)^2 = (b+c)^2& a^2 & a^2 b^2 & (c+a)^2 & b^2 c^2 & c^2 & (a+b)^2 Applying C_1 C_1- C_3, C_2 C_2- C_3, we get = = (b+c)^2-a^2& 0 & a^2 0 & (c+a)^2-b^2 & b^2 c^2-(a+b)^2 & c^2-(a+b)^2 & (a+b)^2 =(a+b+c)^2 b+c-a& 0 & a^2 0 & c+a-b & b^2 c -a - b & c-a- b& (a+b)^2 Applying R_3 R_3- (R_1+R_2), we get =(a+b+c)^2 b+c-a& 0 & a^2 0 & c+a-b & b^2 - 2 b&- 2 a& 2 ab Applying C_1 aC_1 and C_2 bC_2 we get = (a+b+c) a b a b+a c-a^2& 0&a^2 0 & b (c + a-b)&b^2 - 2 b a&- 2 a b&2a b Applying C_1 C_1+ C_3, C_2 _2 + C_3 we get = (a+b+c)^3 a b a (b+c)&a^2&a^2 b^2 & b (a + c)&b^2 0& 0&2 a b = (a+b+c) a b b 2a b b+c&a&a b & c + a&b 0& 0&1 = 2 ab (a + b + c)^2 [(b + c) (c + a) – a b] = 2 ab (a + b + c)^2 [bc + c^2 + a b + a c – a b] = 2 abc (a + b + c)^3

Using properties of determinants, show the following: (b+c)^2& ab…

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Using properties of determinants, show the following: $\begin{vmatrix} (\text{b}+\text{c})^2& \text{ab} & \text{ca} \\ \text{ab} & (\text{b}+\text{c})^2 & \text{bc} \\ \text{ac} & \text{bc} & \text{(a+b)}^2 \end{vmatrix} =2\text{abc}\ (\text{a+b+c})^3\dot{}$

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$\triangle=\begin{vmatrix} (\text{b}+\text{c})^2& \text{ab} & \text{ca} \\ \text{ab} & (\text{b}+\text{c})^2 & \text{bc} \\ \text{ac} & \text{bc} & \text{(a+b)}^2 \end{vmatrix} ; \text{Applying R}_1\rightarrow\text{a R}_1,\text{ R}_2\rightarrow\text{b R}_2,\text{ R}_3\rightarrow\text{c R}_3 \text{ we get}$$=\frac{1}{\text{a b c}}\ \begin{vmatrix} \text{a}(\text{b}+\text{c})^2& \text{b a}^2 & \text{c a}^2 \\ \text{a b}^2 & \text{b}(\text{c}+\text{a})^2 & \text{c b}^2 \\ \text{a c}^2 & \text{b c}^2 &\text{c} \text{(a+b)}^2 \end{vmatrix} $
$=\begin{vmatrix} (\text{b}+\text{c})^2& \text{a}^2 & \text{a}^2 \\ \text{b}^2 & (\text{c}+\text{a})^2 & \text{b}^2 \\ \text{c}^2 & \text{c}^2 & \text{(a+b)}^2 \end{vmatrix} \\ \text{Applying}\text{ C}_1\rightarrow\text{ C}_1-\text{ C}_3,\ \text{C}_2\rightarrow\text{ C}_2-\text{ C}_3, \text{we get}$
$=\triangle=\begin{vmatrix} (\text{b}+\text{c})^2-\text{a}^2& \text{0} & \text{a}^2 \\ \text{0} & (\text{c}+\text{a})^2-\text{b}^2 & \text{b}^2 \\ \text{c}^2-(\text{a+b})^2 & \text{c}^2-\text{(a+b)}^2 & \text{(a+b)}^2 \end{vmatrix}$
$=\text{(a+b+c)}^2\begin{vmatrix} \text{b}+\text{c}-\text{a}& \text{0} & \text{a}^2 \\ \text{0} & \text{c}+\text{a}-\text{b} & \text{b}^2 \\ \text{c }-\text{a} - \text{b} & \text{c}-\text{a}- \text{b}& \text{(a+b)}^2 \end{vmatrix}$
$\text{Applying}\text{ R}_3\rightarrow\text{ R}_3-\text{ (R}_1+\text{R}_2), \text{ we get}$
$ \triangle=\text{(a+b+c)}^2\begin{vmatrix} \text{b}+\text{c}-\text{a}& \text{0} & \text{a}^2 \\ \text{0} & \text{c}+\text{a}-\text{b} & \text{b}^2 \\ -\text{ 2 b}&- \text{ 2 a}& 2\text{ ab} \end{vmatrix}$
$\text{Applying}\text{ C}_1\rightarrow\text{ aC}_1\ \text{and}\text{ C}_2\rightarrow\text{ bC}_2 \text{we get}$
$\triangle=\frac{\text{(a+b+c)}}{\text{a b}}\begin{vmatrix} \text{a b}+\text{a c}-\text{a}^2& \text{0}&\text{a}^2\\ \text{0} & \text{b }(\text{c + a}-\text{b})&\text{b}^2\\ -\text{ 2 b a}&- \text{ 2 a b}&2\text{a b} \end{vmatrix}$
$\text{Applying}\text{ C}_1\rightarrow\text{ C}_1+\text{ C}_3,\text{ C}_2\rightarrow\text{C}_2\ +\ \text{C}_3\ \text{we get}$
$\triangle=\frac{\text{(a+b+c)}^3}{\text{a b}}\begin{vmatrix} \text{a }\text{(b+c)}&\text{a}^2&\text{a}^2\\ \text{b}^2 & \text{b }(\text{a + c})&\text{b}^2\\ \text{0}& \text{0}&\text{2 a b} \end{vmatrix}$
$=\frac{\text{(a+b+c)}}{\text{a b}}\times\text {a b}\times\text {2a b}\begin{vmatrix} \text{b+c}&\text{a}&\text{a}\\ \text{b} & \text{c + a}&\text{b}\\ \text{0}& \text{0}&\text{1} \end{vmatrix}$
$= 2 \text{ab } \text{(a + b + c)}^2 \ [\text{(b + c) (c + a) – a b}]$
$= 2 \text{ ab (a + b + c)}^2 \ \text{[bc + c}^2\text{ + a b + a c – a b]}$
$\text{= 2 abc (a + b + c)}^3$$\dot{}$

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