How do you factor #100 - 81t^6#?

2 Answers
Jan 21, 2017

Answer:

# (10 - 9t^3)(10+9t^3)#

Explanation:

You can recognize squares: 100 is 10^2, 81 is 9^2
Then #100 - 81t^6 = 10^2 - (9t^3)^2 = (10 - 9t^3)(10+9t^3) #

Jan 22, 2017

Answer:

#100-81t^6#

#= (10-9t^3)(10+9t^3)#

#= (root(3)(10)-root(3)(9)t)(root(3)(100)+root(3)(90)t+root(3)(81)t^2)(root(3)(10)+root(3)(9)t)(root(3)(100)-root(3)(90)t+root(3)(81)t^2)#

Explanation:

The difference of squares identity can be written:

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

The difference of cubes identity can be written:

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

The sum of cubes identity can be written:

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

Note also that:

#root(3)(a)root(3)(b) = root(3)(ab)#

Hence we find:

#100-81t^6#

#= 10^2-(9t^3)^2#

#= (10-9t^3)(10+9t^3)#

#= ((root(3)(10))^3-(root(3)(9)t)^3)((root(3)(10))^3+(root(3)(9)t)^3)#

#= (root(3)(10)-root(3)(9)t)(root(3)(100)+root(3)(90)t+root(3)(81)t^2)(root(3)(10)+root(3)(9)t)(root(3)(100)-root(3)(90)t+root(3)(81)t^2)#