Question 622302
There may be a simpler way for this but here goes...<br>
If
{{{a+b+c = 0}}}
then
{{{(a+b+c)^3 = 0}}}
To cube the left side I am going to use the {{{(x+y)^3 = x^3+3x^2y+3xy^2+y^3}}} pattern with the "x" being "(a+b)" and the "y" being "c":
{{{(a+b)^3 + 3(a+b)^2*c + 3(a+b)*c^2 +c^3 = 0}}}
For the {{{(a+b)^3}}} I will use this pattern again:
{{{a^3 + 3a^2b +3ab^2 +b^3 + 3(a+b)^2*c + 3(a+b)*c^2 +c^3 = 0}}}
For the {{{(a+b)^2}}} I will use the {{{(x+y)^2 = x^2+2xy+y^2}}} pattern:
{{{a^3 + 3a^2b +3ab^2 +b^3 + 3(a^2+2ab+b^2)*c + 3(a+b)*c^2 +c^3 = 0}}}
Multiplying we get:
{{{a^3 + 3a^2b +3ab^2 +b^3 + 3a^2*c+6abc+3b^2*c + 3ac^2+3bc^2 +c^3 = 0}}}<br>
Next I'll isolate the cubed terms on the left side by subtracting all the non-cubed terms from both sides:
{{{a^3 +b^3 +c^3 = -3a^2b -3ab^2 - 3a^2*c-6abc-3b^2*c - 3ac^2-3bc^2}}}
Factoring out -3 on the right side:
{{{a^3 +b^3 +c^3 = -3(a^2b +ab^2 +a^2*c+2abc+b^2*c +ac^2+bc^2)}}}
Dividing both sides by abc. (We're told that a b and c are all non-zero so it is OK to divide by this.)
{{{(a^3 +b^3 +c^3)/abc = -3((a^2b +ab^2 +a^2*c+2abc+b^2*c +ac^2+bc^2)/abc)}}}
Splitting up the fractions:
{{{a^3/abc +b^3/abc +c^3/abc = -3(a^2b/abc +ab^2/abc +a^2*c/abc+2abc/abc+b^2*c/abc +ac^2/abc+bc^2/abc)}}}
Simplifying the fractions:
{{{a^2/bc +b^2/ac +c^2/ab = -3(a/c +b/c +a/b+2+b/a +c/b+c/a)}}}
(Note that the left side is the expression we hope is equal to 3.) Adding the like fractions on the right side:
{{{a^2/bc +b^2/ac +c^2/ab = -3((a+b)/c +(a+c)/b+2+ (b+c)/a)}}}
From a + b + c = 0:<ul><li>Subtracting c from each side we get: a + b = -c</li><li>Subtracting b from each side we get: a + c = -b</li><li>Subtracting a from each side we get: b + c = -a</li></ul>Substituting these expressions in for the numerators on the right side we get:
{{{a^2/bc +b^2/ac +c^2/ab = -3((-c)/c +(-b)/b+2+ (-a)/a)}}}
Simplifying the fractions further:
{{{a^2/bc +b^2/ac +c^2/ab = -3(-1 + (-1)+ 2+ (-1))}}}
Simplifying:
{{{a^2/bc +b^2/ac +c^2/ab = -3(-1)}}}
{{{a^2/bc +b^2/ac +c^2/ab = 3}}}