If the sum of the cubes of a and b is 8 and a^6 b^6 = 14, what is

Question

If the sum of the cubes of a and b is 8 and a^6 – b^6 = 14, what is the value of a^3 – b^3?

A. 1/4 
B. 1/2
C. 5/4 
D. 7/4 
E. 2

manpreetsingh86 · Accepted Answer

a^3 + b^3 =8 ---------------1)

(a^3)^2-(b^3)^2=14

we know that a^2-b^2 = (a-b)(a+b)
thus
(a^3+b^3)(a^3-b^3) =14
using 1, we have

a^3 - b^3 = 14/8
=7/4

pairakesh10 · Answer

Give that  a^3 + b^3 = 8 
and 
 a^6 - b^6 = 14

Solve left hand side:
 (a^3)^2 - (b^3)^2 = 14

This is in form  a^2-b^2 . Hence:  (a^3+b^3) (a^3-b^3) = 14

8 (a^3-b^3) = 14

(a^3-b^3)=  14/8 = 7/4

Answer is D

Ashishmathew01081987 · Answer

Given :
 a^3 + b^3 = 8  and  a^6 - b^6 = 14

a^6 - b^6 = (a^3)^2 - (b^3)^2 
 a^6 - b^6 = (a^3 - b^3) (a^3 + b^3)

14 = (a^3 - b^3) 8

(a^3 - b^3) = 7/4

Answer is D

Nunuboy1994 · Answer

this is just another a sub b question-

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

(a^3+b^3)(a^3-b^3)= a^6-b^6= 14

8(a^3-b^3)= 14

(a^3-b^3) = 14/8

d

ScottTargetTestPrep · Answer

We know that a^3 + b^3 = 8

We also have that (a^3 + b^3)(a^3 - b^3) = 14. Substituting, we have:

(8)(a^3 - b^3) = 14

a^3 - b^3 = 14/8 = 7/4

Answer D

ajaynandakumar7 · Answer

Since it is given that a^3+b^3 is 8 and solving for a^6-b^6 which is basically ((a^3)^2-(b^3)^2). 
Implies
14/8 = 7/4

architkap · Answer

can we also solve this by dividing both equations?

a^3+b^3 = 8 1)

a^6-b^6 = 14 2)

Can I just divide 2) by 1)?

a^6/a^3 = a^3
-b^6/b^3 = -b^3
14/8

so we finally get a^3-b^3=14/8 which is the same answer as all above.

egmat · Answer

Hi architkap,

Good instinct to look for a shortcut, and you did land on the right number. But I want to flag something important: the  method you used isn't actually a legal move , and it only gave the right answer here by luck.

Here's the trap. When you "divide term by term," you're really claiming:

(a^6 - b^6) / (a^3 + b^3) = a^6/a^3 - b^6/b^3

That is, you split a fraction across a  sum in the denominator . You can only split a fraction when the denominator is  common  to each piece - like (X + Y)/D = X/D + Y/D. You  cannot  split when the denominator itself is a sum.

Quick proof it fails:  take plain numbers.

- Real value: (6 - 2)/(3 + 1) = 4/4 =  1 
- Term-by-term: 6/3 - 2/1 = 2 - 2 =  0

Different answers - so the move is not valid.

Why it happened to work this time

Your answer came out right only because a^6 - b^6 has a special structure - it's a  difference of squares  in a^3 and b^3:

a^6 - b^6 = (a^3 + b^3)(a^3 - b^3)

So when you divide by a^3 + b^3, that factor cleanly cancels and leaves a^3 - b^3. The  factoring  is what's doing the real work - not the term-by-term division. They coincided here, but on a problem without that clean factor, term-by-term would burn you.

Reliable path:  always factor first.

- a^6 - b^6 = (a^3 + b^3)(a^3 - b^3)
- Sub in the givens:  14  =  8  x (a^3 - b^3)
- So a^3 - b^3 =  14 / 8  =  7/4  -  D

Same answer, but now it's built on a rule you can trust every time.

Answer: D

If the sum of the cubes of a and b is 8 and a^6 b^6 = 14, what is

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If the sum of the cubes of a and b is 8 and a^6 b^6 = 14, what is

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