Question

If \(a + b + c = 11\) and \(ab + bc + ca = 20\), then the value of \(a^3 + b^3 + c^3 - 3abc\) is

• A. 671
• B. 341
• C. 121
• D. 781

Hint:

Use the identity for the sum of cubes and square identities to simplify the expression.

Answer 1

The Correct Option is C

Using the identity: \[\na^3 + b^3 + c^3 - 3abc = (a + b + c)\left(a^2 + b^2 + c^2 - ab - bc - ca\right)\n\] Given \(a + b + c = 11\) and \(ab + bc + ca = 20\), we determine: \[\na^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) = 11^2 - 2 \cdot 20 = 121 - 40 = 81\n\] Therefore: \[\na^3 + b^3 + c^3 - 3abc = 11 \times (81 - 20) = 11 \times 61 = 671\n\] The answer is 671.