Mark the tick against the correct answer in the following:
To find: Value of
We have,
Applying
⇒
Taking (a - b) common
⇒
Applying
⇒
Taking (c-a) common
⇒
Expanding along R1
= (b – a)(c – a)[0 – 1(1 – (a – b)) + (b + c)(0)]
= (b – a)(c – a)(-1 + a – b)
= (b – a)(c – a)(a – b – 1)
= (b – a)(ac – bc – c – a2 + ab + a)
= (abc – b2c – bc – a2b + ab2 + ab – a2c + abc + ac + a3 + a2b + a2)
= 4abc