Question

prove that
| a-b-c 2a 2a |
| 2b b-c-a 2b | =(a+b+c)³
| 2c 2c c-a-b |

Answer 1

Answer:

(a+b+c)³

Step-by-step explanation:

Given matrix is

$\left [\begin {array}{ccc}a-b-c&2a&2a\\2b&b-c-a&2b\\2c&2c&c-a-b \end{array} \right]$

Now, we solve the matrix from the row and column method

= (a-b-c)[(b-c-a)(c-a-b)-4bc] - 2a[2b(c-a-b)-4bc]+2a[4bc-(b-c-a)2c]

= $(a-b-c)[bc-ab-b^{2}-c^{2}+ac+bc-ac-a^{2}+ab-4bc] - 2a [2bc-2ab-2b^{2} -4bc]+ 2a[4bc-2bc+2c^{2}+2ac ]\\= a^{3} +b^{3} +c^{3} + 3a^{2}b +3b^{2}a +3b^{2}c +3c^{2}b +3c^{2}a +3a^{2}c +6abc$

This is the formula of $(a+b+c)^{3}$

Hence, it is proved.

Answer 2

$\left[\begin{array}{ccc}a - b-c&2a&2a\\2b&b-c-a&2b\\2c&2c&c-a-b\end{array}\right]= (a+b+c)^{3}$, hence, it is proved.

Step-by-step explanation:

$L.H.S. =\left[\begin{array}{ccc}a - b-c&2a&2a\\2b&b-c-a&2b\\2c&2c&c-a-b\end{array}\right]$

$Applying, R_{1} : R_{1} + R_{2} +R_{3}$

$=\left[\begin{array}{ccc}a + b+c&a + b+c&a + b+c\\2b&b-c-a&2b\\2c&2c&c-a-b\end{array}\right]$

$Taking common as (a +b +c) inR_{1}$

$=(a+b+c)\left[\begin{array}{ccc}1&1&1\\2b&b-c-a&2b\\2c&2c&c-a-b\end{array}\right]$

Applying, $C_{1} :C_{1} -C_{2}  and C_{2} :C_{2} -C_{3}$

$=(a+b+c)\left[\begin{array}{ccc}0&0&1\\b+c+a&-b-c-a&2b\\0&a+b+c&c-a-b\end{array}\right]$

$Taking common as (a +b + c) inC_{1} and C_{1}$

$=(a+b+c)^{3} \left[\begin{array}{ccc}0&0&1\\1&-1&2b\\0&1&c-a-b\end{array}\right]$

$=(a+b+c)^{3}{0 - 0 + 1 -0}= (a+b+c)^{3} =R.H.S.,Proved$