Problem:
Evaluate the product
( 5 + 6 + 7 ) ( 5 + 6 β 7 ) ( 5 β 6 + 7 ) ( β 5 + 6 + 7 ) . (\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6}-\sqrt{7})(\sqrt{5}-\sqrt{6}+\sqrt{7})(-\sqrt{5}+\sqrt{6}+\sqrt{7}).
( 5 β + 6 β + 7 β ) ( 5 β + 6 β β 7 β ) ( 5 β β 6 β + 7 β ) ( β 5 β + 6 β + 7 β ) .
Solution:
Repeated use of the identity ( x + y ) ( x β y ) = x 2 β y 2 (x+y)(x-y)=x^{2}-y^{2}( x + y ) ( x β y ) = x 2 β y 2 leads to ( 5 + 6 + 7 ) ( 5 + 6 β 7 ) = ( 5 + 6 ) 2 β ( 7 ) 2 = ( 11 + 2 30 ) β 7 = 4 + 2 30 (\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6}-\sqrt{7})=(\sqrt{5}+\sqrt{6})^{2}-(\sqrt{7})^{2}=(11+2 \sqrt{30})-7=4+2 \sqrt{30}( 5 β + 6 β + 7 β ) ( 5 β + 6 β β 7 β ) = ( 5 β + 6 β ) 2 β ( 7 β ) 2 = ( 1 1 + 2 3 0 β ) β 7 = 4 + 2 3 0 β , ( 5 β 6 + 7 ) ( β 5 + 6 + 7 ) = ( 7 ) 2 β ( 5 β 6 ) 2 = 7 β ( 11 β 2 30 ) = β 4 + 2 30 (\sqrt{5}-\sqrt{6}+\sqrt{7})(-\sqrt{5}+\sqrt{6}+\sqrt{7})=(\sqrt{7})^{2}-(\sqrt{5}-\sqrt{6})^{2}=7-(11-2 \sqrt{30})=-4+2 \sqrt{30}( 5 β β 6 β + 7 β ) ( β 5 β + 6 β + 7 β ) = ( 7 β ) 2 β ( 5 β β 6 β ) 2 = 7 β ( 1 1 β 2 3 0 β ) = β 4 + 2 3 0 β and ( 4 + 2 30 ) ( β 4 + 2 30 ) = ( 2 30 ) 2 β 4 2 = 120 β 16 = 104 (4+2 \sqrt{30})(-4+2 \sqrt{30})=(2 \sqrt{30})^{2}-4^{2}=120-16=\boxed{104}( 4 + 2 3 0 β ) ( β 4 + 2 3 0 β ) = ( 2 3 0 β ) 2 β 4 2 = 1 2 0 β 1 6 = 1 0 4 β .
The problems and solutions on this page are the property of the MAA's American Mathematics Competitions