Question 1001892
<pre>
{{{(x-a)^2 + (x-b)^2 + (x-c)^2 + x^2}}} &#8799; {{{a^2 + b^2 + c^2}}}

Two expressions are equal if they differ by zero.  So the procedure 
we will use is to show that the difference between the two expressions 
above that we want to show equal, is zero.

So we examine this difference:

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

Remove the outer parentheses:

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

Arrange them so as to have three differences of squares:

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

Group the three differences of squares:

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

Factor the three differences of squares:

{{{((x-a)^""-a)((x-a)^""+a) +  ((x-b)^""-b)((x-b)^""+b) + ((x-c)^""-c)((x-c)^""+c)+x^2}}}

Remove the inner parentheses:

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

Collect like terms:

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

Distribute to remove the parentheses:

{{{x^2-2ax +  x^2-2bx + x^2-2cx+x^2}}}

Combine like terms:

{{{4x^2-2ax -2bx -2cx}}}

Factor out 2x

{{{2x(2x-a-b-c)}}}

Substitute a+b+c for 2x since that is given:

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

The second parenthetical expression becomes 0:

{{{(a+b+c)(0)}}}

{{{0}}}

The two original expressions differ by 0 so they are equal.

Edwin</pre>