How do you factor #-49 + y^6#?

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Answer 1 · 2015-05-31T00:02:17

#y^6 - 49 = (y^3 - 7)(y^3 + 7)#

Answer 2 · 2015-05-31T00:31:53

#y^6-49=(y^3+7)(y^3-7)#

Problem: Factor #-49+y^6#.

Rewrite the equation as #y^6-49#.

This is an example of the difference of squares:

#(a^2-b^2)=(a+b)(a-b)#.

#a=y^3#
#b=7#

#(y^3)^2-7^2=(y^3+7)(y^3-7)#

Answer 3 · 2015-05-31T09:04:24

#-49+y^6 = y^6-49 = (y^3)^2-7^2 = (y^3 - 7)(y^3 + 7)#

This is as far as you can go with rational coefficients.

If you are allowed real coefficients:

#y^3-7 = y^3-(root(3)(7))^3#

#= (y - root(3)(7))(y^2+(root(3)(7))y+(root(3)(7))^2)#

and

#y^3+7 = y^3+(root(3)(7))^3#

#= (y + root(3)(7))(y^2-(root(3)(7))y+(root(3)(7))^2)#

For shorthand I will write #rho = root(3)(7)#

#-49+y^6#

#= (y - rho)(y^2+rho y+rho^2)(y + rho)(y^2-rho y+rho^2)#