roots, finding angle (1 Viewer)

[lyounamu]

Prove the 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)

How do I work these ones out?

Am I meant to just simplify LHS & RHS. And they will equal?



C) If 2x =a+b+c, show that (x-a)^2+(x-b)^2+(x-c)^2+x^2=a^2+b^2+c^2

Thanks in advance.

A) This is "prove" question. So, you don't need to start from L.H.S. You can right separately and say in the ends that they are same like the followings:

L.H.S = (a+b+c)(ab+bc+ca) - abc
= a^2b + abc + ca^2 + ab^2+b^2c+abc + abc+bc^2+c^2a - abc
= a^2b + abc + ca^2 + ab^2 + b^2c + abc + bc^2 +ac^2

R.H.S = (a+b)(b+c)(c+a)
= (ab + ac + b^2 +bc)(c+a)
= abc + ac^2 + b^2c +bc^2 + a^2b + ca^2 +ab^2 + abc

Therefore, L.H.S = R.H.S

B) L.H.S = (ax + by)^2 + (ay-bx)^2 + c^2(x^2+y^2)
= a^2x^2 + b^2y^2 + 2abxy + a^2y^2 + b^2x^2 -2abxy +c^2x^2 + c^2y^2
= a^2x^2 + b^2y^2 + a^2y^2 + b^2x^2 +c^2x^2 + c^2y^2
= x^2 ( a^2 + b^2 + c^2) + y^2 (a^2 + b^2 + c^2)
= (x^2 + b^2) (a^2 + b^2 + c^2)
= R.H.S

C) L.H.S = (x-a)^2+(x-b)^2+(x-c)^2+x^2
= x^2 - 2ax + a^2 +x^2 - 2bx + b^2 + x^2 - 2cx + c^2 + x^2
= 4x^2 - 2ax - 2bx - 2cx +a^2 + b^2 + c^2
= (2x)^2 - 2x(a +b + c) +a^2 + b^2 +c^2
= 2x (2x - a - b - c ) + a^2 + b^2 + c^2
= 2x (a+b+c - a - b - c ) + a^2 + b^2 + c^2 (since 2x =a+b+c)
= 2x . 0 + a^2 + b^2 +c^2
= a^2 + b^2 + c^2
= R.H.S

You just needed to simplify all thes questions to solve them. When you get the "prove" question, you don't need to start from left-hand side. You can start from both sides and come down to say that they are equal or you can just start from right-hand side.

However, you need to start from left-hand side when you get the "show" questions.