If \(a=12, b=-8\) and \(c=-4\), then find the value of \(a^3+b^3+c^3\)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1152
Add the three numbers first. By the identity \(a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)\), when \(a+b+c=0\), \(a^{3}+b^{3}+c^{3}=3abc\), and no cubes need to be worked out.
Sum of the numbers: \(a+b+c=12-8-4=0\)
So the identity applies: \(a^{3}+b^{3}+c^{3}=3abc=3\times12\times(-8)\times(-4)=1152\)