Let a = 2.3, b = 15.9 and c 3.7. Then [a3+b3+c3 -3abc]/[a2+b2+c2-ab-bc-ca] … (1) As we know that – a3+b3+c3 -3abc = (a+b+c) [ a2+b2+c2-ab-bc-ca] Put the value in Eq (1). = (a+b+c) [ a2+b2+c2-ab-bc-ca] / [ a2+b2+c2-ab-bc-ca] The term [ a2+b2+c2-ab-bc-ca] cancel out and leave – =(a+b+c) =2.3 +15.9 +3.7 =21.9.