Question 1068250
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Factor the following expression completely: {{{ 4a^2c^2-(a^2-b^2+c^2)^2 }}}
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<pre>
  {{{ 4a^2c^2 - (a^2-b^2+c^2)^2 }}} = {{{(2ac)^2 - (a^2-b^2+c^2)^2}}} =     (apply the formula  {{{x^2 - y^2}}} = (x+y)*(x-y) )

= {{{(2ac + (a^2 - b^2 + c^2))*(2ac - (a^2 - b^2 + c^2))}}} = 

= {{{( (a^2 + 2ac + c^2) - b^2)*(b^2 - (a^2 - 2ac + c^2))}}} = 

= {{{(b^2 - (a+c)^2)*(b^2 - (a-c)^2)}}} =                     ( apply again the same formula to EACH of the two factors )

= (b-a-c)*(b+a+c))*(b+a-c)*(b-a+c).


QED.
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