Problem:
Find the value of (52+643β)3/2β(52β643β)3/2.
Solution:
We split 52 into two parts to obtain squares in each set of parentheses:
β(52+643β)3/2β(52β643β)3/2=(43+643β+9)3/2β(43β643β+9)3/2=[(43β+3)2]3/2β[(43ββ3)2]3/2=(43β+3)3β(43ββ3)3=(4343β+3β
3β
43+3β
3243β+33)β(4343ββ3β
3β
43+3β
3243ββ33)=828β
OR
Let Ξ±=(52+643β)1/2 and Ξ²=(52β643β)1/2. We wish to find Ξ±3βΞ²3=(Ξ±βΞ²)(Ξ±2+Ξ±Ξ²+Ξ²2). Now
Ξ±2+Ξ²2=104 and Ξ±Ξ²=(522β36β
43)1/2=(1156)1/2=34
Thus (Ξ±βΞ²)2=Ξ±2β2Ξ±Ξ²+Ξ²2=104β68=36, so Ξ±βΞ²=6 and Ξ±3βΞ²3= 6(104+34)=828β.
The problems on this page are the property of the MAA's American Mathematics Competitions