Question

Prove that $\left|\begin{array}{ccc}a& a+b& a+b+c\\ 2a& 3a+2b& 4a+3b+2c\\ 3a& 6a+3b& 10a+6b+3c\end{array}\right|=a^3$.

Answer 1

$L.H.S.$

$\left|\begin{array}{ccc}a& a+b& a+b+c\\ 2a& 3a+2b& 4a+3b+2c\\ 3a& 6a+3b& 10a+6b+3c\end{array}\right|$

On expand with $C_1$

$a[(3a+2b)(10a+6b+3c)-(6a+3b)(4a+3b+2c)]-2a[(a+b)(10a+6b+3c)-(6a+3b)(a+b+c)]-3a[(a+b)(4a+3b+2c)-(3a+2b)(a+b+c)]$

$\Rightarrow a[30a^2+18ab+9ac+20ab+12b^2+6bc-24a^2-18ab-12ac-12ab-9b^2-6bc]-2a[10a^2+10ab+6ab+6b^2+3ac+3bc-6a^2-6ab-6ac-3ab-3b^2-3bc]-3a[4a^2+3ab+2ac+4ab+3b^2+2bc-3a^2-3ab-3ac-2ab-2b^2-2bc]$

$\Rightarrow a[6a^2+3b^2+8ab-3ac]-2a[4a^2+3b^2+7ab-3ac]-3a[a^2+b^2+2ab-ac]$

$\Rightarrow 6a^3+3a{b}^2+8a^{2}b-3a^{2}c-8a^3-6a{b}^2+14a^{2}b+6a^{2}c-3a^3-3a^{2}b-6a^{2}b+3a^{2}c$

$\Rightarrow a^3$

$R.H.S.$

Hence, proved.