How do you factor #64 - c^12#?
1 Answer
Use some special identities to find:
#64-c^12=(sqrt(2)-c)(sqrt(2)+c)(2+c^2)(2+sqrt(2)c+c^2)(2-sqrt(2)c+c^2)(2+sqrt(6)c+c^2)(2-sqrt(6)c+c^2)#
Explanation:
The difference of cubes identity can be written:
#a^3-b^3=(a-b)(a^2+ab+b^2)#
The difference of squares identity can be written:
#a^2-b^2=(a-b)(a+b)#
Note also that:
#(a^2+ab+b^2)(a^2-ab+b^2) = a^4+a^2b^2+b^4#
The final identity we will use is:
#(a^2+sqrt(3)ab+b^2)(a^2-sqrt(3)ab+b^2) = a^4-a^2b^2+b^4#
These last two identities were found by looking at:
#(a^2+kab+b^2)(a^2-kab+b^2) = a^4+(2-k^2)a^2b^2+b^4#
and picking
Note that
Letting
#64-c^12#
#=4^3-(c^4)^3#
#=(4-c^4)(4^2+4c^4+(c^4)^2)#
#=(4-c^4)(16+4c^4+c^8)#
Next both
#=(2^2-(c^2)^2)(16+4c^4+c^8)#
#=(2-c^2)(2+c^2)(16+4c^4+c^8)#
Next use the difference of squares identity to factor the first quadratic:
#=(sqrt(2)^2-c^2)(2+c^2)(16+4c^4+c^8)#
#=(sqrt(2)-c)(sqrt(2)+c)(2+c^2)(16+4c^4+c^8)#
Next use our special
#=(sqrt(2)-c)(sqrt(2)+c)(2+c^2)(4+2c^2+c^4)(4-2c^2+c^4)#
Then use it again with
#=(sqrt(2)-c)(sqrt(2)+c)(2+c^2)(2+sqrt(2)c+c^2)(2-sqrt(2)c+c^2)(4-2c^2+c^4)#
Then use our final identity with
#=(sqrt(2)-c)(sqrt(2)+c)(2+c^2)(2+sqrt(2)c+c^2)(2-sqrt(2)c+c^2)(2+sqrt(6)c+c^2)(2-sqrt(6)c+c^2)#
That's as far as we can get with Real coefficients.