Difference between revisions of "1963 TMTA High School Algebra I Contest Problem 11"
Coolmath34 (talk | contribs) (Created page with "== Problem == <math>x^3+y^3</math> factored into real primes is <math>\text{(A)} \quad (x+y)(x^2+2xy+y^2) \quad \text{(B)} \quad (x+y)(x^2-2xy+y^2) \quad \text{(C)} \quad (x+...") |
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<math>x^3+y^3</math> factored into real primes is | <math>x^3+y^3</math> factored into real primes is | ||
| − | <math>\text{(A)} \quad (x+y)(x^2+2xy+y^2) \quad \text{(B)} \quad (x+y)(x^2-2xy+y^2) \quad \text{(C)} \quad (x+y)(x^2-xy+y^2) | + | <math>\text{(A)} \quad (x+y)(x^2+2xy+y^2) \quad \text{(B)} \quad (x+y)(x^2-2xy+y^2) \quad \text{(C)} \quad (x+y)(x^2-xy+y^2)</math> |
| − | < | + | <math>\text{(D)} \quad (x+y)(x^2+xy+y^2) \quad \text{(E)} (x+y)(x^2-y^2)</math> |
== Solution == | == Solution == | ||
This is a difference of cubes, so factor using the difference of cubes formula: | This is a difference of cubes, so factor using the difference of cubes formula: | ||
<cmath>x^3 + y^3 = (x+y)(x^2-xy+y^2.</cmath> | <cmath>x^3 + y^3 = (x+y)(x^2-xy+y^2.</cmath> | ||
| − | The answer is < | + | The answer is <math>\boxed{\text{(C)}}.</math> |
| + | |||
== See Also == | == See Also == | ||
{{Succession box | {{Succession box | ||
Revision as of 23:09, 1 February 2021
Problem
factored into real primes is
Solution
This is a difference of cubes, so factor using the difference of cubes formula:
![\[x^3 + y^3 = (x+y)(x^2-xy+y^2.\]](http://latex.artofproblemsolving.com/6/d/b/6db188eaee607186159c02ac73e61b0dbdade78f.png)
The answer is 
See Also
| 1963 TMTA High School Mathematics Contests (Problems) | ||
| Preceded by Problem 10 |
TMTA High School Mathematics Contest Past Problems/Solutions | Followed by Problem 12 |
