Question

$\left[\begin{array}{ccc}b+c& a& a\\ b& c+a& b\\ c& c& a+b\end{array}\right]=4abc;$
Prove that $LHS=RHS$

Answer 1

Given that,

$L.H.S$

$\left[\begin{array}{ccc}b+c& a& a\\ b& c+a& b\\ c& c& a+b\end{array}\right]$

On Expansion of row

$(b+c)\left[\right(a+b\left)\right(c+a)-cb]-a\left[\right(a+b)b-cb]+a[cb-c(c+a\left)\right]$

$\Rightarrow (b+c)[ac+a^2+bc+ab-bc]-a[ab+b^2-bc]+a[cb-c^2-ac]$

$\Rightarrow abc+a^{2}b+a{b}^2+a{c}^2+a^{2}c+abc-a^{2}b-a{b}^2+abc+abc-a{c}^2-a^{2}c$

$\Rightarrow abc+abc+abc+abc$

$\Rightarrow 4abcR.H.S$