Question

| 0 ab² ac² |
| a²b 0 bc² | =2a³b³c³
| a²c b²c 0 |
,prove it using theorems

Answer 1

Answer:

$\left|\begin{array}{ccc}0&ab^2&ac^2\\a^2b&0&bc^2\\a^2c&b^2c&0 \end{array}\right|=2\:a^3b^3c^3$

Step-by-step explanation:

| 0 ab² ac² |

| a²b 0 bc² | =2a³b³c³

| a²c b²c 0 | 

,prove it

$\left|\begin{array}{ccc}0&ab^2&ac^2\\a^2b&0&bc^2\\a^2c&b^2c&0 \end{array}\right|$

$\text{Take a,b and c common from }R_1,\:R_2\:and\:R_3\:\text{ respectively}$

$=abc\:\left|\begin{array}{ccc}0&b^2&c^2\\a^2&0&c^2\\a^2&b^2&0 \end{array}\right|$

$\text{Take }a^2,b^2\:and\:c^2 common from }C_1,\:C_2\:and\:C_3\:\text{ respectively}$

$=(abc)(a^2b^2c^2)\left|\begin{array}{ccc}0&1&1\\1&0&1\\1&1&0 \end{array}\right|$

$=a^3b^3c^3\left|\begin{array}{ccc}0&1&1\\1&0&1\\1&1&0 \end{array}\right|$

$=a^3b^3c^3[0-1(0-1)+1(1-0)]$

$=a^3b^3c^3[0+1+1]$

$=2\:a^3b^3c^3$

$\implies\boxed{\bf\:\left|\begin{array}{ccc}0&ab^2&ac^2\\a^2b&0&bc^2\\a^2c&b^2c&0 \end{array}\right|=2\:a^3b^3c^3}$

Answer 2

Answer:

Hopes it helps...... Mark me as a brainlist