Problem:
Find the value of (52+643​)3/2−(52−643​)3/2.
Solution:
We split 52 into two parts to obtain squares in each set of parentheses:
​(52+643​)3/2−(52−643​)3/2=(43+643​+9)3/2−(43−643​+9)3/2=[(43​+3)2]3/2−[(43​−3)2]3/2=(43​+3)3−(43​−3)3=(4343​+3⋅3⋅43+3⋅3243​+33)−(4343​−3⋅3⋅43+3⋅3243​−33)=828​
OR
Let α=(52+643​)1/2 and β=(52−643​)1/2. We wish to find α3−β3=(α−β)(α2+αβ+β2). Now
α2+β2=104 and αβ=(522−36⋅43)1/2=(1156)1/2=34
Thus (α−β)2=α2−2αβ+β2=104−68=36, so α−β=6 and α3−β3= 6(104+34)=828​.
The problems on this page are the property of the MAA's American Mathematics Competitions