# Question: Eighth Powers

## Comment on Eighth Powers

### ((a+b)^2)^4 = ((a-b)^2)^4

((a+b)^2)^4 = ((a-b)^2)^4

This is a simpler way because in both cases when you expand (a+b)^2 OR (a-b)^2 ; 2ab will be 0. Hence both quantities become (a^2 +b^2)^4.

Great approach!

### But sir am unable to score.

But sir am unable to score. I studied from this website and was able to score 314 on gre. But only 155 in quant :( am giving it again on 30th nov. so going through all the problems in the website again. Any particular advice that you would give to raise the score by at least 5 points in quant ?

### Your score breakdowns don't

Your score breakdowns don't provide much information regarding what's needed to increase your scores.
The GRE is a test of your math and verbal skills AND it's a test of your test-taking skills. So, taking several practice tests is an important part of your prep. This will help you build your test-taking skills, and it will help you identify any remaining area(s) of weakness.
While analyzing your practice tests, there are four main types of weakness to watch out for:
1. specific Quant skills/concepts (e.g., algebra, geometry, etc.)
2. specific Verbal skills/concepts (e.g., vocabulary, 3-blank text completion questions, etc.)
3. test-taking skills (time management, endurance, anxiety etc.)
4. silly mistakes
For the first two weaknesses, the fix is pretty straightforward. Learn the concept/skill and find some practice questions to strengthen that weakness.
If your test-taking skills are holding you back, then you need to work on these. For example, we have a free video about test anxiety at http://www.greenlighttestprep.com/module/general-gre-info-and-strategies...
Finally, if silly mistakes are hurting your score, then it's important that you identify and categorize these mistakes so that, during tests, you can easily spot situations in which you're prone to making errors. I write about this and other strategies in the following article: http://www.greenlighttestprep.com/articles/avoiding-silly-misteaks-gre
Cheers,
Brent

### if we square root 3 times

if we square root 3 times both Quantity A and B, then A: a+b and B:a-b;

Then Quantity A and B would not be equal. Is there any error, if I use this procedure.

In this case, answer is D.

### There's an inherent problem

There's an inherent problem with finding the roots of expressions that are already raised to a power. The reason for this is that EVEN powers will turn any value into a POSITIVE number.

For example, 3^4 = 81 AND (-3)^4 = 81
Likewise, 5^2 = 25 AND (-5)^2 = 25

Now if we find the square root of 5^2 (aka 25), we get 5.
Likewise, if we find the square root of (-5)^2 (aka 25), we also get 5 (not -5).

Many students make the mistake of concluding that √(k²) = k, but this is ONLY true if k is greater than or equal to zero. If k is negative, then √(k²) ≠ k.

For example, √[(-3)²] = √9 = 3 (not -3)

So, with your solution, we cannot say that √(a-b) = a-b, if it's the case that a-b has a NEGATIVE value.

### I took the 8th root to get

I took the 8th root to get rid of the radical sign. Why can't I do that?

### A property of exponents is

A property of exponents is that a number raised to an EVEN power will yield a value that is greater than or equal to zero.

For example, (-2)^8 = 256 AND 2^8 = 256

So, if we know that x^8 = y^8, we can't conclude that x = y. It could be the case that x = -2 and y = 2.

### what if we have (a+b)^5 and

what if we have (a+b)^5 and (a-b)^5

on that case, are we allowed to take 5th root on both side?

### Yes. If the exponent is odd,

Yes. If the exponent is odd, then we can do that.

So, if (a+b)^5 = (a-b)^5, then we can conclude that a + b = a - b

### The question doesn't make

The question doesn't make sense whatsoever. You certainly can take the 8th root of both sides to get a+b for A and a-b for B. You'll find that the answer is D. Question is wrong.

### Rather than just stating the

Rather than just stating the correct answer is D (which is not true), try identifying specific values of a and b that PROVE the correct answer is D. You'll find that it's impossible to do.

Here's a quick example that contradicts your statements that we can simply take the 8th root of both quantities.
QUANTITY A: (-1)^8
QUANTITY B: 1^8
Here it's obvious that the two quantities are equal.

However, look what happens if we take the 8th root of both quantities, as you suggest.
We get:
QUANTITY A: -1
QUANTITY B: 1
By taking the 8th root of both quantities, we incorrectly come to the conclusion that Quantity B is greater.

### Would it be too simplistic to

Would it be too simplistic to deduce that since we know that the power is an EVEN number, quantity B will yield a positive result ie - (a-b)^8 =(a+b) hence answer C?

### That theory will work, but

That theory will work, but only when one of the values is 0 (as in the question).

If neither value is 0, then we can't use that approach.
For example, if a = 2 and b = 1, then:
(a-b)^8 = (2-1)^8 = 1^8 = 1
(a+b)^8 = (2+1)^8 = 3^8 = 6561

Cheers,
Brent

### We can't square root or

We can't square root or perform any other exponent operations in QC problems right? For example, we can't raise both sides to the power 1/8 since we can only perform add, subtract, multiple, and divide matching operations.

### The issue with raising both

The issue with raising both quantities by some power (e.g., power of 1/8) is that the sign (positive or negative) of the base may not be preserved.

For example, we can't conclude that (x^8)^(1/8) = x

Here's what I mean: if x = -1, then (x^8)^(1/8) = 1 (not -1)

Does that help?

Cheers,
Brent

### My mistake here was to think

My mistake here was to think that (-b)^8 is equal to
-(b^8). Parenthesis are really important.This is the type of mistakes I make.

### You're not alone. I often see

You're not alone. I often see students make that same mistake.

### I took this approach: (a+b)^2

I took this approach: (a+b)^2 * (a+b)^2 * (a+b)^2 * (a+b)^2 and expanding (a+b)^2 gives a^2 + b^2 + 2ab, since ab=0, we will be left with (a^2 + b^2)^8 on both sides, therefore answer is C

### That's a solid approach.

That's a solid approach.
I would add that the same property holds true for (a-b)^8, in that (a-b)^2 gives a^2 + b^2 - 2ab, and since ab=0, we get a^2 + b^2.