Question

Show that (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0

Answer 1

`\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(c+a\right)=0`

`\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(c+a\right)-\left(b-c\right)\left(b+c\right)=0-\left(b-c\right)\left(b+c\right)`

`\left(a-b\right)\left(a+b\right)+\left(c-a\right)\left(c+a\right)=-\left(b-c\right)\left(b+c\right)`

`\left(a-b\right)\left(a+b\right)+\left(c-a\right)\left(c+a\right) = a^2-b^2+\left(c-a\right)\left(c+a\right) =a^2-b^2+c^2-a^2 =-b^2+c^2`

`-\left(b-c\right)\left(b+c\right) =-\left(b^2-c^2\right) =-b^2-\left(-c^2\right) =-b^2+c^2`

`-b^2+c^2=-b^2+c^2`

`0=0`

It is always true because 0=0.

Answer 2

• this equation is always true

Explanation:

Show that (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0

(a−b)⋅(a+b)+(b−c)⋅(b+c)+(−a+c)⋅(a+c)=0

(a^2−b^2)+(b^2−c^2)+(−a^2+c^2)=0

(a^2−c^2)+(−a^2+c^2)=0

0=0

this equation is always true