Difference between revisions of "1963 TMTA High School Algebra I Contest Problem 40"
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<math>\text{(D)} \quad 16x^{2}-8xy+4y^{2} \quad \text{(E)} \quad \text{none of these}</math> | <math>\text{(D)} \quad 16x^{2}-8xy+4y^{2} \quad \text{(E)} \quad \text{none of these}</math> | ||
+ | |||
+ | == Solution == | ||
+ | |||
+ | We can factor <math>64x^{3}-8y^{3}</math> as | ||
+ | <cmath>(4x)^{3}-(2y)^{3}=(4x-2y)(16x^{2}+8xy+4y^{2})</cmath> | ||
+ | by using difference of cubes. | ||
+ | |||
+ | Then <math>\frac{64x^{3}-8y^{3}}{4x-2y}</math> is equal to | ||
+ | <cmath>\frac{(4x-2y)(16x^{2}+8xy+4y^{2})}{4x-2y} = 16x^{2}+8xy+4y^{2},</cmath> | ||
+ | or <math>\boxed{\text{(C)}}</math>. | ||
+ | |||
+ | == See Also == | ||
+ | {{Succession box | ||
+ | |header=[[1963 TMTA High School Mathematics Contests]] ([[1963 TMTA High School Algebra I Contest Problems|Problems]]) | ||
+ | |before=[[1963 TMTA High School Algebra I Contest Problem 39 | Problem 39]] | ||
+ | |title=[[TMTA High School Mathematics Contest Past Problems/Solutions]] | ||
+ | |after=Last Problem | ||
+ | }} |
Latest revision as of 12:42, 2 February 2021
Problem
If is divided by , the quotient will be:
Solution
We can factor as by using difference of cubes.
Then is equal to or .
See Also
1963 TMTA High School Mathematics Contests (Problems) | ||
Preceded by Problem 39 |
TMTA High School Mathematics Contest Past Problems/Solutions | Followed by Last Problem |