(b%2Bc)(c%2Ba)%2Babc-factorise.jpeg "(a+b)(b+c)(c+a)+abc Factorise")
Factorization of (a+b)(b+c)(c+a) + abc
In this article, we will explore the factorization of the expression (a+b)(b+c)(c+a) + abc. This expression may seem complex at first, but with some clever algebraic manipulation, we can simplify it and reveal a beautiful underlying structure.
Step 1: Expand the Expression
Let's start by expanding the expression (a+b)(b+c)(c+a) using the distributive property of multiplication over addition:
$(a+b)(b+c)(c+a) = (ab + ac + bb + bc)(c+a)$ $= abc + a^2c + abc + ac^2 + bbca + bc^2 + acb + a^2b$ $= 2abc + a^2c + ac^2 + bbca + bc^2 + acb + a^2b$
Step 2: Add abc to the Expression
Now, let's add abc to the expanded expression:
$2abc + a^2c + ac^2 + bbca + bc^2 + acb + a^2b + abc$ $= 3abc + a^2c + ac^2 + bbca + bc^2 + acb + a^2b$
Step 3: Factorize the Expression
Observe that the expression has a common factor of abc. We can factor it out:
$3abc + a^2c + ac^2 + bbca + bc^2 + acb + a^2b$ $= abc(3 + a/b + c/a + b/c + c/b + a/c + b/a)$ $= abc(3 + a/b + c/a + b/c + c/b + a/c + b/a)$ $= abc((a+b+c)/a + (a+b+c)/b + (a+b+c)/c)$ $= abc((a+b+c)(1/a + 1/b + 1/c))$ $= (a+b+c)(1/a + 1/b + 1/c)abc$
And there you have it! The expression (a+b)(b+c)(c+a) + abc factorizes beautifully into (a+b+c)(1/a + 1/b + 1/c)abc. This factorization reveals the underlying symmetry of the expression and can be used to simplify calculations and prove identities in various mathematical contexts.
