Subtract the sum of 3l−4m−7n2 and 2l+3m−4n2 from the sum of 9l+2m−3n2 and −3l+m+4n2........
Sum of 9l+2m−3n2 and −3l+m+4n2=9l+2m−3n2+(−3l)+m+4n2=9l+2m−3n2−3l+m+4n2=9l−3l+2m+m−3n2+4n2=6l+3m+n2and sum of 3l−4m−7n2 and 2l+3m−4n2=3l−4m−7n2+2l+3m−4n2=3l+2l−4m+3m−7n2−4n2=5l−m−11n2Now (6l+3m+n2)=(5l−m−11n2)=6l+3m+n2−5l+m+11n2=6l−5l+3m+m+n2+11n2=l+4m+12n2