Difference between revisions of "2020 AMC 10A Problems/Problem 14"
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Continuing to combine <cmath>\frac{x^3+y^3}{x^2}+\frac{x^3+y^3}{y^2}=\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}=\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}</cmath> | Continuing to combine <cmath>\frac{x^3+y^3}{x^2}+\frac{x^3+y^3}{y^2}=\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}=\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}</cmath> | ||
From the givens, it can be concluded that <math>x^2y^2=4</math>. Also, <cmath>(x+y)^2=x^2+2xy+y^2=16</cmath> This means that <math>x^2+y^2=20</math>. Substituting this information into <math>\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}</math>, we have <math>\frac{(20)(4)(22)}{4}=20\cdot 22=\boxed{\textbf{(D)}\ 440}</math>. | From the givens, it can be concluded that <math>x^2y^2=4</math>. Also, <cmath>(x+y)^2=x^2+2xy+y^2=16</cmath> This means that <math>x^2+y^2=20</math>. Substituting this information into <math>\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}</math>, we have <math>\frac{(20)(4)(22)}{4}=20\cdot 22=\boxed{\textbf{(D)}\ 440}</math>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/ZGwAasE32Y4 | ||
| + | |||
| + | ~IceMatrix | ||
==See Also== | ==See Also== | ||
Revision as of 22:18, 31 January 2020
Real numbers 
and 
satisfy 
and 
. What is the value of![\[x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y?\]](http://latex.artofproblemsolving.com/1/b/7/1b77f92edbab9a6134e14217ee0c56700cef3dbf.png)
Solution
![\[x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y=x+\frac{x^3}{y^2}+y+\frac{y^3}{x^2}=\frac{x^3}{x^2}+\frac{y^3}{x^2}+\frac{y^3}{y^2}+\frac{x^3}{y^2}\]](http://latex.artofproblemsolving.com/3/c/b/3cb720eb6b236c7897b32cbe34768bfa695b2ba3.png)
Continuing to combine ![\[\frac{x^3+y^3}{x^2}+\frac{x^3+y^3}{y^2}=\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}=\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}\]](http://latex.artofproblemsolving.com/7/4/4/74411ff57b4c958d41ca7cb6c6d0230b331b0ad7.png)
From the givens, it can be concluded that 
. Also, ![\[(x+y)^2=x^2+2xy+y^2=16\]](http://latex.artofproblemsolving.com/f/e/4/fe4d79118d31b37765c0ae83dc534680c73f2ee4.png)
This means that 
. Substituting this information into 
, we have 
.
Video Solution
~IceMatrix
See Also
| 2020 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
