Prove that:

Solution:

Let LHS $=\Delta=\mid \begin{array}{ll}1 & \mathrm{a}^{2}+\mathrm{bc} & \mathrm{a}^{3}\end{array}$

$1 \quad b^{2}+c a \quad b^{3}$

$\begin{array}{lcc}1 & c^{2}+a b & c^{3}\end{array}$

$\Rightarrow \Delta=\mid 0 \quad\left(a^{2}+b c\right)-\left(b^{2}+c a\right) \quad a^{3}-b^{3}$

$0 \quad\left(b^{2}+c a\right)-\left(c^{2}+a b\right) \quad b^{3}-c^{3}$

$\begin{array}{llll}1 & \mathrm{c}^{2}+\mathrm{ab} & \mathrm{c} & {\left[\text { Applying } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2} \quad \text { and } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\right]}\end{array}$

$=\mid \begin{array}{lll}0 & & a^{2}-b^{2}-c a+b c & a^{3}-b^{3}\end{array}$

$\begin{array}{lll}0 & b^{2}-c^{2}-a b+c a & b^{3}-c^{3} \\ 1 & c^{2}+a b & c^{3}\end{array}$

1 $c^{2}+a b$ $c^{3} \mid$

$=\mid \begin{array}{llllll}0 & (\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b}-\mathrm{c}) & (\mathrm{a}-\mathrm{b})\left(\mathrm{a}^{2}+\mathrm{ab}+\mathrm{b}^{2}\right) 0 & (\mathrm{~b}-\mathrm{c})(\mathrm{b}+\mathrm{c}-\mathrm{a}) & (\mathrm{b}-\mathrm{c})\left(\mathrm{b}^{2}+\mathrm{bc}+\mathrm{a}^{2}\right) 1 & \mathrm{c}^{2}+\mathrm{ab}\end{array}$

$\begin{array}{llll}=(a-b)(b-c) \mid 0 & a+b-c & a^{2}+a b+b^{2} \\ 0 & (b+c-a) & \left(b^{2}+b c+c^{2}\right) & \end{array}$

$\begin{array}{cccc}1 & c^{2}+a b & c^{3} \mid & {\left[\text { Taking out }(a-b) \text { common from } R_{1} \text { and }(b-c) \text { from } R_{2}\right]}\end{array}$

$=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c}) \mid 0 \quad \mathrm{a}+\mathrm{b}-\mathrm{c} \quad \mathrm{a}^{2}+\mathrm{ab}+\mathrm{b}^{2} 0 \quad(\mathrm{~b}+\mathrm{c}-\mathrm{a})-(\mathrm{a}+\mathrm{b}-\mathrm{c}) \quad\left(\mathrm{b}^{2}+\mathrm{bc}+\mathrm{c}^{2}\right)-\left(\mathrm{a}^{2}+\mathrm{ab}+\mathrm{b}^{2}\right) 1$

[Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}$ ]

$\begin{array}{llllll} & =(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c}) \mid 0 & \mathrm{a}+\mathrm{b}-\mathrm{c} & \mathrm{a}^{2}+\mathrm{ab}+\mathrm{b}^{2} 0 & 2(\mathrm{c}-\mathrm{a}) & \mathrm{b}(\mathrm{c}-\mathrm{a})+\left(\mathrm{c}^{2}-\mathrm{a}^{2}\right) 1 \\ & =(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a}) \mid 0 & \mathrm{a}+\mathrm{b}-\mathrm{c} & \mathrm{a}^{2}+\mathrm{ab}+\mathrm{b}^{2} & \end{array}$

$\begin{array}{ccc}0 & 2 & \mathrm{a}+\mathrm{b}+\mathrm{c} \\ 1 & \mathrm{c}^{2}+\mathrm{ab} & \mathrm{c}^{3}\end{array}$

$=(a-b)(b-c)(c-a) \times\left\{1 \times\left|a+b-c \quad a^{2}+a b+b^{2} \quad 2 \quad a+b+c\right|\right\} \quad\left[\right.$ Expanding along $\left.C_{1}\right]$

$=(a-b)(b-c)(c-a) \times\left\{(a+b)^{2}-c^{2}-\left(2 a^{2}+2 a b+2 b^{2}\right)\right\}$

$=(a-b)(b-c)(c-a)\left\{(a+b)^{2}-c^{2}-(a+b)^{2}-\left(a^{2}+b^{2}\right)\right\}$

$=-(a-b)(b-c)(c-a)\left(a^{2}+b^{2}+c^{2}\right)$

$=R H S$

Hence proved.