Question 1119606
Factoring:
a)16b^2c^2-4(b^2+c^2-a^2)

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b)(x^2+4y^2-20)^2-16(xy-4)^2
Same here.  Some "simplification" might be possible after using the difference of squares, but Step 1 is the difference of squares.
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{{{(x^2+4y^2-20)^2-16(xy-4)^2 = ((x^2+4y^2-20)-4(xy-4))*((x^2+4y^2-20)+4(xy-4))}}}
= {{{((x^2+4y^2-4)-4xy)*((x^2+4y^2-36)+4xy)}}}
= {{{(x^2+4y^2-4-4xy)*(x^2+4y^2-36+4xy)}}}
= {{{(x^2-4xy+4y^2-4)*(x^2+4xy+4y^2-36)}}}
Now each of the 4nomials can be factored.
= {{{((x-2y)^2-4)*((x+2y)^2-36)}}}
= {{{(x-2y-2)(x-2y+2)*(x+2y-6)*(x+2y+6)}}}
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Other than knowing about the difference of squares, it's just careful bookkeeping.
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Look up difference of 2 cubes, also.