If \(a+b+c=6\) and \(a b+b c+c a=10\), then the value of \(a^3+b^3+c^3-3 a b c\) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 36
The expression is the left side of the identity \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\), so only \(a^2+b^2+c^2\) has to be found from the two given sums.
Sum of the squares: \(a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=36-20=16\)
Second bracket of the identity: \(a^2+b^2+c^2-(ab+bc+ca)=16-10=6\)
Value of the expression, the first bracket times the second: \(6\times6=36\)
Tip: The same identity can be written as \((a+b+c)\left[(a+b+c)^2-3(ab+bc+ca)\right]\), which uses the two given sums directly: \(6\times(36-30)=36\).