How do you factor #64 - c^12#?

Question

962 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-15T00:13:55

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)#

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 #k = 1# or #k = sqrt(3)# to get the required result.

Note that #64 = 4^3# and #c^12 = (c^4)^3# are both perfect cubes. So we can use the difference of cubes identity to start to factorise this.

Letting #a=4# and #b=c^4# we find:

#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 #4=2^2# and #c^4 = (c^2)^2# are perfect squares so we can factor some more:

#=(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 #a^4+a^2b^2+b^4# identity with #a=2# and #b=c^2# to factor some more:

#=(sqrt(2)-c)(sqrt(2)+c)(2+c^2)(4+2c^2+c^4)(4-2c^2+c^4)#

Then use it again with #a=sqrt(2)# and #b=c# to factor some more:

#=(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 #a=sqrt(2)# and #b=c# to factor the last quartic:

#=(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.