Prove that :
(i) 
= (a − b) (b − c) (c − a) (a + b + c)
(ii) 
= − (a + b + c) (a − b) (b − c) (c − a)
(iii) 
= 4 abc
(iv)If 
= (a + b) (b + c) (c + a) 
.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i) 
= (a − b) (b − c) (c − a) (a + b + c)
Applying C 1 → C 1 – C 2 & C 2 → C 2 – C 3
= (a – b) (b – c) 
= (a – b) (b – c) (c – a) (a + b + c)
(ii) 
Applying C 2 → C 2 + C 1
= (a + b + c) 
= – (a + b + c) 
= – (a + b + c) (a – b) (b – c) (c – a)
(iii) 
by R 1 → R 1 + R 2 + R 3
= 2 
by R 1 → R 1 – R 3
= 2 
= 2 [b((c + a)(a + b) – cb) – a(ab + b 2 – bc)]
= 2 [b(ac + cb + a 2 + ab – cb) – a 2 b – ab 2 + abc]
= 2 [abc + a 2 b + ab 2 – a 2 b – ab 2 + abc] = 4 abc
(iv) 
= K 
⇒ (a 2 – b 2 ) (b 2 – c 2 ) (c 2 – a 2 ) = K(a – b) (b – c) (c – a)
∴ K = (a + b) (b + c) (c + a)
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