Question

Question for Brainly Teachers​

Answer 1

Answer:

see solution in attachment bro ,

very nice question but lengthy ...

mark as brainlist if you understand ...

Answer 2

Step-by-step explanation:

$\left|\begin{array}{ccc}</p><p>a-b-c&2b&2c\\2a&b-c-a&2c\\2a&2b&c-a-b\end{array}\right|=(a+b+c)^3$

C1->C1+C2+C3

$\left|\begin{array}{ccc}</p><p>a-b-c+2b+2c&2b&2c\\2a+b-c-a+2c&b-c-a&2c\\2a+2b-c-a-b&2b&c-a-b\end{array}\right|$

$\left|\begin{array}{ccc}</p><p>a+b+c&2b&2c\\a+b+c&b-c-a&2c\\a+b+c&2b&c-a-b\end{array}\right|$

take (a+b+c) common from C1

$(a+b+c)\left|\begin{array}{ccc}</p><p>1&2b&2c\\1&b-c-a&2c\\1&2b&c-a-b\end{array}\right|$

R1->R1-R3

R2->R2-R3

$(a+b+c)\left|\begin{array}{ccc}</p><p>1-1&2b-b+c+a&2c-2c\\1-1&b-c-a-2b&2c-c+a+b\\1&2b&c-a-b\end{array}\right|$

$(a+b+c)\left|\begin{array}{ccc}</p><p>0&a+b+c&0\\0&-(a+b+c)&a+b+c\\1&2b&c-a-b\end{array}\right|$

take common (a+b+c) from R1 and R2

$(a+b+c)^3\left|\begin{array}{ccc}</p><p>0&1&0\\0&-1&1\\1&2b&c-a-b\end{array}\right|$

Expand the determinant along C1

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

So,

$\left|\begin{array}{ccc}</p><p>a-b-c&2b&2c\\2a&b-c-a&2c\\2a&2b&c-a-b\end{array}\right|=(a+b+c)^3$

Hence Proved.

Hope it helps you.