Difference between revisions of "1975 AHSME Problems/Problem 12"
Quantummech (talk | contribs) (Created page with "We can factor <math>a^3-b^3=19x^3</math> into: <cmath>(a-b)(a^2+ab+b^2)=19x^3.</cmath> Substituting yields: <cmath>x(a^2+ab+b^2)=19x^3</cmath> <cmath>a^2+ab+b^2=19x^2.</cmath>...") |
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| + | == Problem 12== | ||
| + | If <math>a \neq b, a^3 - b^3 = 19x^3</math>, and <math>a-b = x</math>, which of the following conclusions is correct? | ||
| + | |||
| + | <math>\textbf{(A)}\ a=3x \qquad \textbf{(B)}\ a=3x \text{ or } a = -2x \qquad \textbf{(C)}\ a=-3x \text{ or } a = 2x \qquad \\ \textbf{(D)}\ a=3x \text{ or } a=2x \qquad \textbf{(E)}\ a=2x</math> | ||
| + | |||
| + | ==Solution== | ||
We can factor <math>a^3-b^3=19x^3</math> into: | We can factor <math>a^3-b^3=19x^3</math> into: | ||
<cmath>(a-b)(a^2+ab+b^2)=19x^3.</cmath> | <cmath>(a-b)(a^2+ab+b^2)=19x^3.</cmath> | ||
Revision as of 05:59, 17 May 2016
Problem 12
If 
, and 
, which of the following conclusions is correct?
Solution
We can factor 
into:
![\[(a-b)(a^2+ab+b^2)=19x^3.\]](http://latex.artofproblemsolving.com/f/7/2/f7221a757b3e6ef14225cffa9688471165f91bfe.png)
Substituting yields:
![\[x(a^2+ab+b^2)=19x^3\]](http://latex.artofproblemsolving.com/e/c/7/ec7386a2b7981da9ca99059f0e8c042fae1ae1c8.png)
![\[a^2+ab+b^2=19x^2.\]](http://latex.artofproblemsolving.com/2/3/c/23c370a1db3a7b4658351e45c8a397a999c19a94.png)
This is equal to:
![\[(a-b)^2+3ab=19x^2\]](http://latex.artofproblemsolving.com/2/a/f/2afd58794b8609608087f6cdb20e5a0dfb14cfd6.png)
![\[x^2+3ab=19x^2\]](http://latex.artofproblemsolving.com/d/e/d/dedc682661a286d8de93680ba2c11ccd61d229f9.png)
![\[ab=6x^2.\]](http://latex.artofproblemsolving.com/3/e/b/3ebd5e6de4892a1488b3b619501d997e74fcd8ed.png)
Checking with the possible answers, along with 
yields the only answer to be \box{(B): 
or 
}.
