How do you factor 100 - 81t^6100−81t6?
2 Answers
Explanation:
You can recognize squares: 100 is 10^2, 81 is 9^2
Then
100-81t^6100−81t6
= (10-9t^3)(10+9t^3)=(10−9t3)(10+9t3)
= (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)=(3√10−3√9t)(3√100+3√90t+3√81t2)(3√10+3√9t)(3√100−3√90t+3√81t2)
Explanation:
The difference of squares identity can be written:
a^2-b^2 = (a-b)(a+b)a2−b2=(a−b)(a+b)
The difference of cubes identity can be written:
a^3-b^3 = (a-b)(a^2+ab+b^2)a3−b3=(a−b)(a2+ab+b2)
The sum of cubes identity can be written:
a^3+b^3=(a+b)(a^2-ab+b^2)a3+b3=(a+b)(a2−ab+b2)
Note also that:
root(3)(a)root(3)(b) = root(3)(ab)3√a3√b=3√ab
Hence we find:
100-81t^6100−81t6
= 10^2-(9t^3)^2=102−(9t3)2
= (10-9t^3)(10+9t^3)=(10−9t3)(10+9t3)
= ((root(3)(10))^3-(root(3)(9)t)^3)((root(3)(10))^3+(root(3)(9)t)^3)=((3√10)3−(3√9t)3)((3√10)3+(3√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)=(3√10−3√9t)(3√100+3√90t+3√81t2)(3√10+3√9t)(3√100−3√90t+3√81t2)
