SOLUTION: Prove these identities. a) (a+b+c)(ab+bc+ca)−abc= (a+b)(b+c)(c+a) b) (ax+by)^2 + (ay−bx)^2 +c^2(x^2 +y^2)= (x^2 +y^2)(a^2 +b^2 +c^2) Thank you so much :)

Algebra ->  Equations -> SOLUTION: Prove these identities. a) (a+b+c)(ab+bc+ca)−abc= (a+b)(b+c)(c+a) b) (ax+by)^2 + (ay−bx)^2 +c^2(x^2 +y^2)= (x^2 +y^2)(a^2 +b^2 +c^2) Thank you so much :)      Log On


   



Question 1139327: Prove these identities.
a) (a+b+c)(ab+bc+ca)−abc= (a+b)(b+c)(c+a)
b) (ax+by)^2 + (ay−bx)^2 +c^2(x^2 +y^2)= (x^2 +y^2)(a^2 +b^2 +c^2)
Thank you so much :)

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
a)
%28a%2Bb%2Bc%29%28ab%2Bbc%2Bca%29-abc=+%28a%2Bb%29%28b%2Bc%29%28c%2Ba%29
manipulate left side
%28a%2Bb%2Bc%29%28ab%2Bbc%2Bca%29-abc
=
=a%5E2b%2Babc%2Ba%5E2c+++%2Bab%5E2%2Bb%5E2c%2Bbca+++%2Bbc%5E2%2Bc%5E2a+ ...group
=

=a%5E2%28b%2Bc%29%2Bab%28b%2Bc%29++%2Bca%28c%2B+b+++%29+%2B+bc+%28b%2Bc%29++
=%28a%5E2%2Bab++%2Bca+%2B+bc%29+%28b%2Bc%29+
=%28%28a%5E2%2Bab%29++%2B%28ca+%2B+bc%29%29+%28b%2Bc%29+
=%28a%28a%2Bb%29++%2Bc%28a+%2B+b%29%29+%28b%2Bc%29+
=%28a+%2Bc%29%28a+%2B+b%29+%28b%2Bc%29++

b.)

manipulate left side:
%28ax%2Bby%29%5E2+%2B+%28ay-bx%29%5E2+%2Bc%5E2%28x%5E2+%2By%5E2%29
=
=
=x%5E2+%28a%5E2++%2B+b%5E2+%2Bc%5E2%29+%2By%5E2%28a%5E2+%2B+b%5E2++%2Bc%5E2%29
=%28x%5E2++%2By%5E2%29%28a%5E2+%2B+b%5E2++%2Bc%5E2%29