How do you factor #a^8 - a^2b^6#?

1 Answer
May 10, 2016

Answer:

#a^2(a-b)(a^2+ab+b^2)(a+b)(a^2-ab+b^2)#

Explanation:

#a^(2+6) - a^2b^6 #

#= a^2a^6-a^2b^6#

#=a^2(a^6-b^6)#, here #color(red)(a^2)# is common between the terms

#=a^2(a^(3xx2)-b^(3xx2))#

#=a^2[(a^3)^2-(b^3)^2]#

#color(red)("This is of the form " x^2-y^2 = (x-y)(x+y), "where " x=a^3 " and " y = b^3)#

#=a^2(a^3-b^3)(a^3+b^3)#

Now we factorize #(a^3-b^3)# and #(a^3+b^3)#:

#color(red)("We know that " (a^3-b^3)=(a-b)(a^2+ab+b^2)#

#color(red)("And "(a^3+b^3)=(a+b)(a^2-ab+b^2)#

Then,

#a^2(a^3-b^3)(a^3+b^3)=color(blue)(a^2(a-b)(a^2+ab+b^2)(a+b)(a^2-ab+b^2)#