What is the value of a^4 - b^4 ?

Bunuel,

A question here, why can't we assume that since a+b=8 then a^2+b^2=64? same as a^2-b^2=16, then sq root both sides and get a-b=4

Thank you!

Let me try to answer.

If you are given \(a+b=8\), then you CAN NOT write it as \(a^2+b^2=64\). The reason is that you must square both sides COMPLETELY,

i.e.

\(a+b=8\) ---> squaring both sides ---> \((a+b)^2=8^2\) ---> \(a^2+2ab+b^2=64\)

Think of it this way, if I tell you that a+b=8 ----> I can have 1 case for (a,b) as a=7, b=1 , does \(a^2+b^2=64\)? NO.

Also, I dont know where you saw that then \(a^2-b^2=16\) --->\(a-b=4\) but this is again incorrect.

If you have \(a^2-b^2=16\) ---> the only correct manipulation for this expression is \((a+b)(a-b)=16\) and nothing else.

Again, if \(a^2-b^2=16\), I can have \(a= \sqrt{17}\) and \(b=1\), now does a-b = 4 ? NO

The formulae that I have used above are (learn these as they are one of the most used formulae in GMAT Quant):

\((a + b) = a^2 + 2ab + b^2\)

\((a - b) = a^2 - 2ab + b^2\)

\(a^2-b^2 = (a+b)*(a-b)\)

FYI,

1. If you are given a+b = 8 , then a^2+b^2=64 ONLY IF 2ab = 0 ---> ab=0 ---> either a or b=0 or even BOTH a,b =0. Thus the only case when if a+b = 8 , then a^2+b^2=64, is when either a OR b =0.

2. If you are given a^2-b^2 = 16 ---> a-b=4 , ONLY POSSIBLE when b=0 and a=4



Hope this helps.

Generally \((a+b)^2\neq a^2+b^2\) and \((a-b)^2 \neq a^2-b^2\). This is algebra 101. You need to brush up basics before attempting questions.

Please go through the following links:

ALL YOU NEED FOR QUANT ! ! !: all-you-need-for-quant-140445.html

Theory on Algebra: algebra-101576.html
Algebra - Tips and hints: algebra-tips-and-hints-175003.html

Factoring Quadratics: https://www.purplemath.com/modules/factquad.htm
Solving Quadratic Equations: https://www.purplemath.com/modules/solvquad.htm

DS Algebra Questions to practice: search.php?search_id=tag&tag_id=29
PS Algebra Questions to practice: search.php?search_id=tag&tag_id=50

Special algebra set: new-algebra-set-149349.html

Hope this helps.

Great explanations, actually now I get the special cases much better

Yeah you are right, I am still discovering "what can be done and what can't be done" in algebra.

Thank you for the quick reply!

Hello,

Can we not take the first equation a^2-b^2= 16 as straight away as (a+b) (a-b)=16. Clearly a^2 has to be Greater than b^2
To satisfy this, a has to be 5 and b has to be 3 . Eventually 5^4- 3^4 can be easily found.(625-81). so only the first should suffice

Hi qazi11

It is not necessary that both a and b are integers

Imagine a case when a=\(\sqrt{18}\) and b=\(\sqrt{2}\), here \(a^2 - b^2 = 16\)
Now, you will have a different value when you find \(a^4 - b^4\)

Hope this helps you!

Can I use the following:

(a^2-b^2= (a+b) (a-b))
(a+b) (a-b)=16
(a+b)=8 and a-b=2
which gives the value of a as 5 and b as 3??
Doesn't it make 1 sufficient alone?

Hi ritikajain1988

from the equation (a+b)(a-b)=16, you cannot assume that a+b=8 & a-b=2, it could also be a+b=16 & a-b=1 or a+b=-8 & a-b=-2 etc.

Essentially you have two variable and one equation, hence it cannot be solved

Hello

You have taken a+b = 8, but that is given in the second statement. Since you have taken data from second statement and used it in first statement, the answer becomes C here.

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

(1) \(a^2 - b^2 = 16\)

We don't know what a^2 + b^2 equals. Insufficient.

(2) \(a + b = 8\)

Clearly insufficient.

(1&2) \((a+b)(a-b) = 16\)

\(8(a-b) = 16\)

\(a - b = 2\)

\(2a = 10\)

\(a = 5\)

\(b = 3\)

\(a^2 + b^2 = 25 + 9 = 34\)

\(34 * 16 \)

SUFFICIENT.

Answer is C.

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