Difference between revisions of "1963 TMTA High School Algebra I Contest Problem 40"

Revision as of 12:02, 2 February 2021

If $64x^{3}-8y^{3}$ is divided by $4x-2y$ , the quotient will be:

$\text{(A)} \quad 4x^{2}-2y^{2} \quad \text{(B)} \quad 4x^{2}+2y^{2} \quad \text{(C)} \quad 16x^{2}+8xy+4y^{2}$

$\text{(D)} \quad 16x^{2}-8xy+4y^{2} \quad \text{(E)} \quad \text{none of these}$

We can factor $64x^{3}-8y^{3}$ as $(4x)^{3}-(2y)^{3}=(4x-2y)(16x^{2}+8xy+4y^{2})$ by using difference of cubes.

Then $\frac{64x^{3}-8y^{3}}{4x-2y}$ is equal to $\frac{(4x-2y)(16x^{2}+8xy+4y^{2})}{4x-2y} = 16x^{2}+8xy+4y^{2},$ or $\boxed{\text{(C)}}$ .

@@ Line 6: / Line 6: @@
 <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>.