Question 1209745
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Fill in the blanks, to complete the factorization:


(a^2 + b^2 - c^2)^2 - 4a^2 b^2 - 4a^2 c^2 + 4b^2 c^2 = (a + ___)(a + ___)(a + ___)(a + ___)
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        Step 1.  Decompose into the product of two quadratic polynomials


    (a^2 + b^2 - c^2)^2 - 4a^2*b^2 - 4a^2*c^2 + 4b^2*c^2 =

  =  a^4 + b^4 + c^4 + 2a^2*b^2 - 2a^2*c^2 - 2b^2*c^2 - 4a^2*b^2 - 4a^2*c^2 + 4b^2*c^2 = 


           next step make a routine combining like terms


  = a^4 + b^4 + c^4 - 2a^2*b^2 - 6a^2*c^2 + 2b^2*c^2 = 


           next step make grouping/re-grouping


  = (a^4 + b^4 + c^4 2 - 2a^2*b^2 - 2a^2*c^2 + 2b^2*c^2) - 4a^2*c*2 = 


           next step complete the squares


   = (-a^2 + b^2 + c^2)^2 - 4a^2*c^2 = 


           next step factor as the difference of squares 


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


           next step is changing the signs everywhere in both parentheses 
           and light re-arranging in each parentheses (for further convenience)


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



        Step 2.  Decompose each parentheses as the product of linear binomials relative "a"



Now we want to decompose first parentheses (a^2 + 2ac - (b^2 + c^2)).    (1)

Consider this aggregate as a standard quadratic trinomial  a^2 + 2ac + X relative to variable 'a'.


Remember how to decompose a trinomial via its roots


     a^2 + 2ac + X = {{{(a - (-c + (sqrt(d))/2))*(a - (-c - (sqrt(d))/2))}}},    (2)


where d is the discriminant.  In this case, the discriminant is


     d = (2c)^2 + 4*(b^2+c^2) = 4c^2 + 4b^2 + 4c^2 = 4(b^2+2c^2).

 
Therefore, decomposition for expression (1) takes the form

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

                          = {{{(a + (c - sqrt(b^2+2c))) * (a + (c + sqrt(b^2+2c^2)))}}}.   (3)




Now, we want to decompose second parentheses (a^2 - 2ac - (b^2 + c^2)).    (4)

By analogy, consider this aggregate as a standard quadratic trinomial  a^2 - 2ac + X.


Remember how to decompose a trinomial via its roots


     a^2 - 2ac + X = {{{(a - (c + (sqrt(d))/2)) * (a + (c - (sqrt(d))/2))}}},    (5)


where d is the discriminant.  In this case, the discriminant is the same


     d = (2c)^2 + 4*(b^2+c^2) = 4c^2 + 4b^2 + 4c^2 = 4(b^2+2c^2).

 
Therefore, decomposition for expression (5) takes the form

    a^2 - 2ac - (b^2+c^2) = {{{(a - (c + sqrt(b^2+2c))) * (a - (c - sqrt(b^2+2c^2)))}}}.    (6)



Combining everything above, we get finally this remarkable decomposition

    (a^2 + b^2 - c^2)^2 - 4a^2 b^2 - 4a^2 c^2 + 4b^2 c^2 = 

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


which is the required form.


So, the four blanks are  {{{(c - sqrt(b^2+2c^2))}}},  {{{(c + sqrt(b^2+2c^2))}}},  {{{-(c + sqrt(b^2+2c^2))}}}  and  {{{-(c - sqrt(b^2+2c^2))}}}.

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Solved.