How do you factor #100 - 81t^6#?
2 Answers
Explanation:
You can recognize squares: 100 is 10^2, 81 is 9^2
Then
#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)#
