Question

Find the sum of
(viii) 𝑎𝑏𝑐+2𝑏𝑎+3𝑎𝑐, 4𝑐𝑎−4𝑎𝑏+2𝑏𝑐𝑎 and 2𝑎𝑏−3𝑎𝑏𝑐−6𝑎𝑐

Answer 1

Step: Find the sum of given expressions
We need to find sum of (𝑎𝑏𝑐+2𝑏𝑎+3𝑎𝑐), (4𝑐𝑎−4𝑎𝑏+2𝑏𝑐𝑎) and (2𝑎𝑏−3𝑎𝑏𝑐−6𝑎𝑐)
= (𝑎𝑏𝑐+2𝑏𝑎+3𝑎𝑐)+(4𝑐𝑎−4𝑎𝑏+2𝑏𝑐𝑎)+(2𝑎𝑏−3𝑎𝑏𝑐−6𝑎𝑐)
=(𝑎𝑏𝑐+2𝑏𝑐𝑎−3𝑎𝑏𝑐)+(2𝑏𝑎−4𝑎𝑏+2𝑎𝑏)+(3𝑎𝑐+4𝑐𝑎−6𝑎𝑐) [Grouping like terms]
=(3𝑎𝑏𝑐−3𝑎𝑏𝑐)+(4𝑎𝑏−4𝑎𝑏)+(7𝑎𝑐−6𝑎𝑐)
=0+0+𝑎𝑐=𝑎𝑐
Hence, the answer is 𝑎𝑐.