Question

If a is not equal to b, and a^3-b^3=19x^3, and a-b=x, which of the following conclusions is correct?
a=3x
a=3x or a=-2x
a=-3x or a=2x
a=3x or a=2x
a=2x

Answer 1

\[a^3-b^3=(a-b)(a^2+ab+b^2)=19x^3\]
\[a-b=x\]
\[a^2+ab+b^2=19x^2\]

Answer 2

i guess that does not solve the problem does it?

Answer 3

I got that far...but I don't know how to go from there.

Answer 4

i don't know lets see what we can try
\[a-b=x\]
\[(a-b)^2=x^2=a^2-ab+b^2\] so we have
\[x^2=a^2-ab+b^2\] and
\[19x^2=a^2+ab+b^2\]

Answer 5

sorry it should be
\[a^2-2ab+b^2=x^2\]

Answer 6

19x^2-x^2=a^2+ab+b^2-(a^2-2ab+b^2)\]
\[18x^2=-3ab\] well that still doesn't do it does it?

Answer 7

i can't seem to get rid of the "b"

Answer 8

try some of the a's given...a=3x or a=-2x work

Answer 9

a=3x or a=-2x

http://www.wolframalpha.com/input/?i=solve+for+a%3A+a%5E3-b%5E3%3D19x%5E3%3B+a-b%3Dx

Answer 10

if a=3x then b=2x
Is there a typo in the solutions given?

Answer 11

\[a^+ab+b^2=19x^2\]
\[b=a-x\]
\[a^2+a(a-x)+a^2-2ax+x^2=19x^2\]
\[a^2+a^2-ax+a^2-2ax+x^2=19x^2\]
3a^2-3ax=18x^2

Answer 12

\[3a^2-3ax-18x^2=0\]

Answer 13

\[a^2-ax-6x^2=0\]

Answer 14

@cathyangs Is there a typo in the solutions given?

Answer 15

\[(a-3x)(a+2x)=0\]

Answer 16

very nice

Answer 17

\[a=3x or -2x\]

Answer 18

Thank you. That really means something coming from The Great One as it were.

Answer 19

jackie gleason was the great one!

Answer 20

So you're in excellent company.