Question

[xa÷xb] a+b÷3 × (x6÷xc) b+6÷3 × (x6÷x9) c+9÷4​

Answer 1

Answer:

Here's the fastest way I know how:

Subtract 3 from both sides but creatively:

x−ab+c−1+x−ba+c−1+x−ca+b−1=0

x−a−b−cb+c+x−a−b−ca+c+x−a−b−ca+b=0

Taking common:

(x−a−b−c)×k=0

Where k is some random expression in a,b, and c.

This implies,

x−a−b−c=0

x=a+b+c