Evaluate (20^5+70^5)/(20^4-5460000+70^4 )

[@als2cents679]

Here is something that is almost purely computational
(20^5 + 70^5) / (20^4 - 5460000 + 70^4)
= (10^5 * (2^5 + 7^5)) / (10^4 * (7^4 - 546 + 2^4))
= 10 * (32 + 7 * (50 - 1)^2) / (49^2 - 2*49*4 + 4^2 + 2*49*4 - 546)
= 10 * (32 + 7 * (2500 - 100 + 1)) / ((49 - 4)^2 + 8 * (50 - 1) -546)
= 10 * (32 + 7 * 2401) / (45^2 + 400 - 8 - 546)
= 10 * (32 + 16807) / (25 * 81 - 154)
= 10 * 16839 / (8100 / 4 - 154)
= 10 * 3 * 5613 / (2025 - 154)
= 30 * 3 * 1871 / 1871
= 90

[@MyOneFiftiethOfADollar]

@@als2cents679 nice way to do it that does not rely on that fifth power algebraic factoring identity used in video.
In math contests, the competitors are sometimes looking for ways to speed up the computation due to the time constraints.

[@MyOneFiftiethOfADollar]

Your method relies on 20^5 and 70^5 having a "nice" common power of 10, right?

[@als2cents679]

@@MyOneFiftiethOfADollar Yeah, just like your solution relies on the fact that both of them are to the power of 5, no?

[@MyOneFiftiethOfADollar]

@@als2cents679 what I meant to convey is any quotient in the form
(x^5 + y^5/(x^4 -C + y^4) = x+y whether x and y have common factors or not.
The way you did it is purely arithmetic.
Try it your way for x=23 and y=74. I am interested in computational efficiency and appreciate your help and interest.

[@als2cents679]

@@MyOneFiftiethOfADollar Yeah, it does not work as well for that case.
What I would have done I guess is similar to what you have, viz. knowing that x^5 + y^5 is divisible by (x + y), do the long division to get the other number, then realize it is same as denominator, to get the answer.