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2006 AMC 10A Problem 2

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2006 AMC 10A Problem 2AMC 10 Step by Step
Name:                             Date: September 13, 2026
Functions · difficulty 1 of 5 · about 1 min · position P1-10

Define x⊗y=x3−yx\otimes y=x^3-y . What is h⊗(h⊗h)h\otimes (h\otimes h) ?

  1. A)

    −h-h

  2. B)

    00

  3. C)

    hh

  4. D)

    2h2h

  5. E)

    h3h^3

Answer

C

Key insight

Evaluate the inner operation first, h ⊗ h = h^3 − h, then feed that result in as the second input of the outer operation.

Solution

Work from the inside out. The rule says "cube the first input and subtract the second," so
h⊗h=h3−h.h \otimes h = h^3 - h.
Now the outer operation has first input hh and second input h3−hh^3 - h:
h⊗(h3−h)=h3−(h3−h)=h3−h3+h=h.h \otimes (h^3 - h) = h^3 - (h^3 - h) = h^3 - h^3 + h = h.

The answer is (C) h\boxed{\textbf{(C)}\ h}.

Why this works

A made-up operation is just a two-variable function; x⊗yx \otimes y is not symmetric here, so which input is "first" matters. Nested expressions are evaluated innermost first, and the result is substituted as a single block, parentheses included.

The trap

Putting the inner result into the first slot or dropping the parentheses, so h^3 − h^3 − h = −h (choice (A)) appears instead of h^3 − (h^3 − h).

Common mistakes

  • Putting the inner result into the first slot or dropping the parentheses, so h3−h3−h=−hh^3 - h^3 - h = -h (choice (A)) appears instead of h3−(h3−h)h^3 - (h^3 - h).
  • Cubing the inner result, (h3−h)3−h(h^3 - h)^3 - h, by mixing up which input gets cubed.

Techniques

Set up the equation/formula and compute; no special trick needed