How do you simplify # [ (a^2 - b^2)^3 + (b^2 - c^2)^3 + (c^2 - a^2)^3 ] / [ (a^4 - b^4)^3 + (b^4 - c^4)^3 + (c^4 - a^4)^3]#?

1 Answer
May 10, 2016

Answer:

#((a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3)/((a^4-b^4)^3+(b^4-c^4)^3+(c^4-a^4)^3)=1/((a^2+b^2)(b^2+c^2)(c^2+a^2))#

excluding any of #a=+-b#, #b=+-c#, #c=+-a#.

Explanation:

Notice that:

#(A-B)^3+(B-C)^3+(C-A)^3#

#=(color(red)(cancel(color(black)(A^3)))-3A^2B+3AB^2-color(red)(cancel(color(black)(B^3))))+(color(red)(cancel(color(black)(B^3)))-3B^2C+3BC^2-color(red)(cancel(color(black)(C^3))))+(color(red)(cancel(color(black)(C^3)))-3C^2A+3CA^2-color(red)(cancel(color(black)(A^3))))#

#=3(AB^2-A^2B+BC^2-B^2C+CA^2-C^2A)#

#=3(A-B)(B-C)(C-A)#

Note also:

#a^4-b^4 = (a^2-b^2)(a^2+b^2)#, etc.

So:

#((a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3)/((a^4-b^4)^3+(b^4-c^4)^3+(c^4-a^4)^3)#

#=(color(red)(cancel(color(black)(3)))(a^2-b^2)(b^2-c^2)(c^2-a^2))/(color(red)(cancel(color(black)(3)))(a^4-b^4)(b^4-c^4)(c^4-a^4))#

#=(color(red)(cancel(color(black)((a^2-b^2))))color(red)(cancel(color(black)((b^2-c^2))))color(red)(cancel(color(black)((c^2-a^2)))))/(color(red)(cancel(color(black)((a^2-b^2))))(a^2+b^2)color(red)(cancel(color(black)((b^2-c^2))))(b^2+c^2)color(red)(cancel(color(black)((c^2-a^2))))(c^2+a^2))#

#=1/((a^2+b^2)(b^2+c^2)(c^2+a^2))#

excluding any of #a=+-b#, #b=+-c#, #c=+-a#