Question

show that: (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0

Answer 1

I hope it helps you ...............................

Answer 2

Given,

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

To prove,

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

Solution,

We can simply solve this mathematical problem using the following process:

As per the algebraic identities;

If x and y are two variables,

then (x+y)(x-y) = x^2 - y^2

Now,

LHS = (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)

= {(a-b)(a+b)} + {(b-c)(b+c)} + {(c-a)(c+a)}

= {(a^2 - b^2)} + {(b^2 - c^2)} + {(c^2 - a^2)}

{Using the algebraic identity: (x+y)(x-y) = x^2 - y^2}

= {a^2 - b^2 + b^2 - c^2 + c^2 - a^2}

= 0 = RHS

Hence, it is proved that (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0.