Solution: Since $ p(x) $ is a cubic polynomial, when dividing by $ x^4 - 1 $, the remainder $ r(x) $ must be a polynomial of degree less than 4. However, because $ x^4 - 1 $ has roots $ \pm 1, \pm i $, and $ p(x) $ is real, the remainder can be expressed as any cubic or lower, but here we are to find the **actual remainder**, which is unique and of degree $ < 4 $. But since $ p(x) $ is already degree 3, and division by degree 4 polynomial yields a unique remainder of degree $ < 4 $, and since $

["Title: Understanding the Remainder When Dividing a Cubic Polynomial by $ x^4 - 1 $", "When dividing a cubic polynomial $ p(x) $ by the quartic polynomial $ x^4 - 1 $, a fundamental principle of polynomial division tells us that the remainder $ r(x) $ must be a polynomial of degree strictly less than 4. But here lies a subtle and important insight: since $ p(x) $ is already a cubic polynomial (and thus degree 3), and the divisor $ x^4 - 1 $ has four distinct roots — $ 1, -1, i, -i $ — we can determine the exact remainder through evaluation and interpolation.", "### The Mathematics Behind the Remainder", "By the polynomial division algorithm, any polynomial $ p(x) $ divided by $ x^4 - 1 $ yields a unique quotient $ q(x) $ and a remainder $ r(x) $ such that:\n$$\np(x) = (x^4 - 1)q(x) + r(x),\n$$\nwhere $ \deg(r(x)) < 4 $. Since $ p(x) $ is cubic, $ \deg(p(x)) = 3 $, which is less than 4 — but this does not mean $ r(x) = p(x) $ necessarily, because the division still reserves space for lower-degree remainder under the constraint of uniqueness.", "However, because $ x^4 - 1 $ vanishes at $ x = 1, -1, i, -i $, evaluating $ p(x) $ at these points gives:\n$$\np(1) = r(1),\quad p(-1) = r(-1),\quad p(i) = r(i),\quad p(-i) = r(-i).\n$$\nThis means the remainder $ r(x) $ is uniquely determined as the uniquely interpolating polynomial of degree $ \leq 3 $ satisfying:\n$$\nr(\pm1) = p(\pm1),\quad r(\pm i) = p(\pm i).\n$$\nThus, even though the remainder must have degree less than 4, it is not $ p(x) $ unless $ p(x) $ fits the degree-3 or lower structure at these 4 points — which it may or may not.", "### Constructing the Remainder", "Since $ p(x) $ is cubic, we can express it as:\n$$\np(x) = a_3x^3 + a_2x^2 + a_1x + a_0,\n$$\nand the remainder $ r(x) $ upon division by $ x^4 - 1 $ is a real polynomial of degree at most 3. Using the values at $ x = 1, -1, i, -i $, we can construct $ r(x) $ via Lagrange interpolation. These evaluations define a continuous function on the complex points $ {1, -1, i, -i} $, and the interpolating cubic polynomial is unique.", "Therefore, the remainder $ r(x) $ is the unique cubic (or lower-degree) polynomial matching $ p(x) $ at these four points — and since $ r(x) $ agrees with $ p(x) $ at the roots of $ x^4 - 1 $, it is fully determined.", "But crucially, because $ p(x) $ is cubic and the divisor is degree 4, the remainder must be of degree $ \leq 3 $, and the evaluation conditions at 4 distinct points fix $ r(x) $ completely. Any remaining ambiguity is resolved by the fact that $ p(x) $ and $ r(x) $ agree at four points, and $ \deg(r) < 4 $, so they must be identical only if $ p(x) $ itself is degree ≤3 — but in general, $ r(x) $ is the best approximation in the Euclidean sense.", "However, since $ p(x) $ is degree 3 and the remainder has degree strictly less than 4, the remainder is simply the interpolating cubic polynomial through the four values $ (1, p(1)), (-1, p(-1)), (i, p(i)), (-i, p(-i)) $. This interpolated $ r(x) $ is the exact remainder, and no further simplification is possible without knowledge of $ p(x) $.", "### Conclusion", "In summary, when dividing a cubic polynomial $ p(x) $ by $ x^4 - 1 $, the remainder $ r(x) $ is uniquely determined — not automatically $ p(x) $, but a degree-≤3 polynomial constructed via interpolation at the four roots of $ x^4 - 1 $. Thus, the remainder is fully specified by the values of $ p(x) $ at $ x = 1, -1, i, -i $, and is the unique cubic or lower polynomial satisfying these constraints.", "This highlights a key insight: while the degree of the remainder is bounded by $ \deg(x^4 - 1) - 1 = 3 $, its exact form depends on $ p(x) $, and division yields a unique, well-defined result — which in this case is the interpolating polynomial matching $ p(x) $ at four strategic points.", "For practical computation, one can use Lagrange interpolation:", "$$\nr(x) = p(1)\frac{(x+1)(x-i)(x+i)}{(1+1)(1-i)(1+i)} + p(-1)\frac{(x-1)(x-i)(x+i)}{(-1-1)(-1-i)(-1+i)} + p(i)\frac{(x-1)(x+1)(x+i)}{(i-1)(i+1)(i+i)} + p(-i)\frac{(x-1)(x+1)(x-i)}{(-i-1)(-i+1)(-i-i)}.\n$$", "Although complex in form, this expression yields a real cubic remainder when $ p(x) $ is real.", "In summary: The remainder upon dividing a cubic $ p(x) $ by $ x^4 - 1 $ is not arbitrary — it is uniquely determined by interpolation at the four roots of the divisor, and is a cubic or lower polynomial that agrees with $ p(x) $ at $ x = 1, -1, i, -i $. This guarantees a definite, computable result, valid for all real cubic polynomials.", "---", "Keywords: cubic polynomial remainder, division by $ x^4 - 1 $, remainder theorem, polynomial interpolation, $ r(x) < 4 $, division algorithm polynomial, $ x = 1, -1, i, -i $, Lagrange interpolation, real polynomial remainder.\nMeta Description:** When dividing a cubic polynomial by $ x^4 - 1 $, the remainder is a degree-3 or lower polynomial uniquely determined by evaluating $ p(x) $ at the four roots of $ x^4 - 1 $. Learn how to compute the exact remainder using interpolation."]