How do you factor 100−81t6?
2 Answers
Jan 21, 2017
Explanation:
You can recognize squares: 100 is 10^2, 81 is 9^2
Then
Jan 22, 2017
100−81t6
=(10−9t3)(10+9t3)
=(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:
a2−b2=(a−b)(a+b)
The difference of cubes identity can be written:
a3−b3=(a−b)(a2+ab+b2)
The sum of cubes identity can be written:
a3+b3=(a+b)(a2−ab+b2)
Note also that:
3√a3√b=3√ab
Hence we find:
100−81t6
=102−(9t3)2
=(10−9t3)(10+9t3)
=((3√10)3−(3√9t)3)((3√10)3+(3√9t)3)
=(3√10−3√9t)(3√100+3√90t+3√81t2)(3√10+3√9t)(3√100−3√90t+3√81t2)
