Maybe I'm approaching my question wrong, but this seems very familiar to one I got on an actual test about a week ago:
5^1/3+95^1/3+207^1/3=x^1/3; what is x?

I had, and still have, absolutely no idea how to solve it.

Please Sir

read the comment above provided by the tutor

Regards

I think we should have much simpler solution, but please let me try.

#STEP1 : simplify numerator.
132^5 - 2(132^4 ) + 6(132^3)-3(132)

Focus on left 2 elements.
132^5 - 2(132 ^ 4 ) =(132^4)(132 - 2) = 132^4(130)
130 is 2 x 65 so, remainder of 132^4(130 ) should be zero, so we can ignore.

#STEP2 calculate right 2 elements of numerator.
This was not fun part me , but..

6(132^3)-3(132) = 132(6 x 132^2 -3 ) = 132(17424 x 6-3) = 130 x (17424 x 6 -3 ) + 2 x (17424 x 6 -3 ) = 130 x 104541 + 2 x 104541
We know that result of above multiplication will be end XX....2 , so we can guess that answer should be B or D.
But I could not determine which was correct , so I needed to calculate rest .

We can ignore 130 x 104541 part, rest is 2 x 104541 = 209082

#STEP3 Calculate remainder.
It's enough simple to calculate for me.
209082 / 65 = 3216 with 42 , so answer is B) , 42

Remainder when 132 is divided by 65 is 2, 4 for 132^2, 8 for 132^5 and so on. So we just need to solve the numerator by replacing 132 with 2.
2^5 - 2(2^4) + 6(2^3) - 3(2) = 42.
I don't know if this approach is mathematically correct or not but it yields the correct answer.

I think your solution seems to be correct, and much simpler, just for one note, I think you wanted to write 8 for 132^3 not 8 for 132^5 ?

We need to find the remainder of \(132^5 − 2*132^4 + 6*132^3 −3*132\) by 65

We solve these problems by using Binomial Theorem, where we split the number into two parts, one part is a multiple of the divisor(65 or 130) and a big number, other part is a small number.

=> \(132^{5}\) = \((130 + 2)^{5}\)

Watch this video to MASTER BINOMIAL Theorem

Now, when we expand this expression then all the terms except the last term will be a multiple of 130 or 65.
=> All terms except the last term will give 0 as remainder then divided by 65
=> Problem is reduced to what is the remainder when the last term is divided by 65
=> What is the remainder when \(5C5 * 130^0 * 2^{5}\) is divided by 65 = Remainder of \(2^{5}\) by 65 = 32

=> Remainder of \(132^5 \) by 65 = remainder of 32 by 65 = 32

Remainder of 2*132^4 by 65 = Remainder of 2*2^4 by 65 = Remainder of 32 by 65 = 32
Remainder of 6*132^3 by 65 = Remainder of 6*2^3 by 65 = Remainder of 48 by 65 = 48
Remainder of 3*132 by 65 = Remainder of 3*(130 + 2) by 65 = Remainder of 3*2 by 65 = 6

=> Total Remainder = 32 - 32 + 48 - 6 = 42

So, Answer will be B
Hope it helps!

Watch following video to MASTER Remainders by 2, 3, 5, 9, 10 and Binomial Theorem