Question

by using properties of determination show that: | a-b-c 2a 2a | | 2b b-c-a 2b | =(a+b+c)³ | 2c 2c c-a-b |​

Answer 1

Answer:

VERIFIED ANSWER

Step-by-step explanation:

REFER THE ATTACHED IMAGE

Answer 2

$\Large\mathtt\green{ }\huge\underline\mathtt\red{Answer : }$

We have,

2b

b-c-a2b

2c2c

C-a-b|

Applying RiR +R2 + R,]

a +b+c a+b+c a+b+c

2bb-c-a

2b2c

2c C-a-b

[Taking (a +b+c) common from the first row]

(a +b+c)2b b-c-a 2c 2c

2bC-a-b

[Applying CC-C and CC-C]

=(a +b+c)0-(a +b+c) 2b a+b+c

a +b+c

C-a-b

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