Maths Olympiad Prep

Track / Stage 3 / 152 of 260 #792 of 2604

Problem 792

AMC 10/12, early questions
Algebra Difficulty 3.8 Find the answer Harvard-MIT Math Tournament · United States

If a@b=a3−b3a−ba @ b = \frac{a^{3} - b^{3}}{a - b}, for how many real values of aa does a@1=0a @ 1 = 0?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Next problem →

Official solution

Solution:

If a3−1a−1=0\frac{a^{3} - 1}{a - 1} = 0, then a3−1=0a^{3} - 1 = 0, or (a−1)(a2+a+1)=0(a - 1)(a^{2} + a + 1) = 0.

Thus a=1a = 1, which is an extraneous solution since that makes the denominator of the original expression 00, or aa is a root of a2+a+1a^{2} + a + 1.

But this quadratic has no real roots, in particular its roots are −1±−32\frac{-1 \pm \sqrt{-3}}{2}.

Therefore there are no such real values of aa, so the answer is 0\mathbf{0}.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.