Answer :
Answer:
[tex]a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)=0[/tex]
Step-by-step explanation:
This is so simple,
Let ' s simplify it out,
Step 1: Using the identity,
[tex]x^2-y^2=(x+y)(x-y)[/tex]
[tex]a(b-c)(b+c)+b(c-a)(c+a)+c(a-b)(a+b)[/tex]
Step 2: Distributing each term,
[tex]a(b-c)(b+c)+b(c-a)(c+a)+c(a-b)(a+b)[/tex]
Step 3: Reordering the terms:
[tex]a(b-c)(b+c)+b(c-a)(c+a)+c(a-b)(a+b)[/tex]
Therefore,
[tex]a(b-c) (b+c)+b(c-a) (c+a)+c(a+b)(a-b) = 0[/tex]
So, the simplified form of the given expression is:
[tex]a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)=0[/tex]