Difference between revisions of "1963 TMTA High School Algebra I Contest Problem 11"

Latest revision as of 23:14, 1 February 2021

$x^3+y^3$ factored into real primes is

$\text{(A)} \quad (x+y)(x^2+2xy+y^2) \quad \text{(B)} \quad (x+y)(x^2-2xy+y^2) \quad \text{(C)} \quad (x+y)(x^2-xy+y^2)$

$\text{(D)} \quad (x+y)(x^2+xy+y^2) \quad \text{(E)} (x+y)(x^2-y^2)$

This is a difference of cubes, so factor using the difference of cubes formula: $x^3 + y^3 = (x+y)(x^2-xy+y^2).$ The answer is $\boxed{\text{(C)}}.$

@@ Line 8: / Line 8: @@
 == Solution ==
 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 <math>\boxed{\text{(C)}}.</math>

1963 TMTA High School Mathematics Contests (Problems)
Preceded by Problem 10	TMTA High School Mathematics Contest Past Problems/Solutions	Followed by Problem 12