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If the sum of three numbers is 18 and the sum of their squares is 36, then find the difference between the sum of their cubes and three times their product.

1. 1449
2. -1944
3. -1494
4. 4149

Correct answer: -1944

Solution

Let the numbers be \(a,b,c\). From \((a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\), we get \(18^2=36+2(ab+bc+ca)\), so \(ab+bc+ca=144\). Using \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\), the value is \(18(36-144)=18(-108)=-1944\).

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