Question

Show that (a^2+1)(c^2+1)=(ac-1)^2+(a+c)^2

Answer 1

See proof below

Explanation:

On the left hand side we have

`\left(a^2+1\right)\left(c^2+1\right)`

Apply the FOIL method to get

`a^2c^2+a^2\cdot \:1+1\cdot \:c^2+1\cdot 1`

Simplify to get

`a^2c^2+a^2+c^2+1`

On the right hand side we have

`\left(ac-1\right)^2+\left(a+c\right)`

Expand

`\left(ac-1\right)^2`

=
`(ac)^2 -2\cdot (ac)(1) + (-1)^2\\\\`

`=a^2c^2-2ac+1` (1)

Expand

`\left(a+c\right)^2`

`=a^2+2ac+c^2` (2)

Add (1) and (2) to get the entire right side

`=a^2c^2-2ac+1 + a^2+2ac+c^2\\\\= a^2c^2 + 1 + a^2 + c^2\\\\\\= a^2c^2 + a^2 + c^2 + 1 \\\\` (by rearranging terms)

Looking at the left side and right side simplified expressions we see they are the same.

Hence proved

Answer 2

Explanation:

`(a^2+1)(c^2+1)=\\\\a^2c^2+c^2+a^2+1=\\\\a^2c^2+(2ac-2ac)+c^2+a^2+1=\\\\(a^2c^2-2ac+1)+(a^2+2ac+c^2)=\\\\(ac-1)^2+(a+c)^2`