\(\boldsymbol{(p-t)(q-t)(r-t)=-f(t)}\)

This identity is a classic algebraic shortcut popular in middle‑ and high‑school competition mathematics (Olympiad algebra), known as the three‑variable cubic universal formula. It connects a monic cubic polynomial with its three roots via a simple sign transformation.

1. Notation Definition

• Let \(p, q, r\) be the three roots (real or complex) of a monic cubic polynomial (leading coefficient = 1):\(f(t)=t^3+at^2+bt+c\)where \(a,b,c\) are constant coefficients.
• t is an arbitrary independent variable or parameter.

2. Rigorous Derivation

By the Factor Theorem in algebra: if \(p,q,r\) are roots of polynomial \(f(t)\), \(f(t)\) can be factorized as:

\(f(t)=(t-p)(t-q)(t-r)\)

Factor out \(-1\) from each binomial term on the right‑hand side:

\(f(t)=(-1)^3(p-t)(q-t)(r-t)=-(p-t)(q-t)(r-t)\)

Rearranging terms directly yields the universal formula:

\(\boldsymbol{(p-t)(q-t)(r-t)=-f(t)}\)

3. Core Mathematical Meaning

The formula means the product of terms \((\text{root} – t)\) for all three roots of a monic cubic polynomial equals the negative value of the polynomial evaluated at t. It avoids tedious expansion of triple brackets and directly converts root‑related products into polynomial evaluation.

4. Reason for the Name “Universal”

It is called “universal” for three key reasons:

1. It works for any monic cubic polynomial, no matter whether its roots are real, irrational or complex;
2. It drastically simplifies calculations of symmetric sums, root‑product expansions and Vieta‑formula‑based problems;
3. It acts as a general‑purpose transformation tool for all cubic‑root algebraic problems.

5. Typical Application

Combined with Vieta’s formulas for cubics:

\(p+q+r=-a,\quad pq+qr+rp=b,\quad pqr=-c\)

we can substitute specific values of t (e.g., \(t=0,1,-1\)) into the formula to quickly compute products like pqr or \((p-1)(q-1)(r-1)\) without lengthy expansion.

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The roots of 

 are 

 and 

 What is the value of

Background of the Universal Formula

For a cubic polynomial f(x)=x3+ax2+bx+c with roots p,q,r, the universal formula is:

(p−t)(q−t)(r−t)=−f(t)

This formula lets us directly compute the product (p−t)(q−t)(r−t) by plugging t into f(x), without expanding long brackets.

Problem Given

f(x)=x3+2×2−x+3, roots p,q,r. Evaluate (p2+4)(q2+4)(r2+4).

Step 1: Factor the quadratic term using complex numbers

Recall the difference‑of‑squares identity for imaginary numbers:

a2+4=a2−(2i)2=(a−2i)(a+2i)

where i=−1​ (imaginary unit).

Thus:

(p2+4)(q2+4)(r2+4)=(p−2i)(p+2i)(q−2i)(q+2i)(r−2i)(r+2i)=[(p−2i)(q−2i)(r−2i)]⋅[(p+2i)(q+2i)(r+2i)]

Step 2: Apply the universal formula twice

• For t=2i: (p−2i)(q−2i)(r−2i)=−f(2i)
• For t=−2i: (p+2i)(q+2i)(r+2i)=−f(−2i)

So the target product becomes:

[−f(2i)]⋅[−f(−2i)]=f(2i)⋅f(−2i)

Step 3: Compute f(2i) and f(−2i)

f(2i)​=(2i)3+2(2i)2−(2i)+3=8i3+2(4i2)−2i+3=−8i−8−2i+3=−5−10i​

f(−2i)​=(−2i)3+2(−2i)2−(−2i)+3=−8i3+2(4i2)+2i+3=8i−8+2i+3=−5+10i​

Step 4: Multiply the two results

f(2i)⋅f(−2i)=(−5−10i)(−5+10i)=(−5)2−(10i)2=25−(−100)=125

Final answer: 125 (Option D)

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中文详细讲解（万能公式核心解题思路）

万能公式回顾

若三次多项式 f(x)=x3+ax2+bx+c 的三个根为 p,q,r，则万能公式：

(p−t)(q−t)(r−t)=−f(t)

核心作用：直接把根的连乘 (p−t)(q−t)(r−t) 转化为多项式代入求值，省去复杂展开。

题目条件

多项式 f(x)=x3+2×2−x+3，根为 p,q,r，求 (p2+4)(q2+4)(r2+4)。

步骤 1：复数因式分解（关键变形）

利用虚数平方差：a2+4=a2−(2i)2=(a−2i)(a+2i)（i=−1​，虚数单位）

原式拆分为：

(p2+4)(q2+4)(r2+4)=[(p−2i)(q−2i)(r−2i)]⋅[(p+2i)(q+2i)(r+2i)]

步骤 2：套用万能公式

把万能公式里的 t 分别换成 2i 和 −2i：

• 令 t=2i：(p−2i)(q−2i)(r−2i)=−f(2i)
• 令 t=−2i：(p+2i)(q+2i)(r+2i)=−f(−2i)

因此原式 = [−f(2i)]⋅[−f(−2i)]=f(2i)⋅f(−2i)

步骤 3：代入计算 f(2i)、f(−2i)

利用 i2=−1, i3=−i：

f(2i)=(2i)3+2(2i)2−2i+3=−8i−8−2i+3=−5−10i

f(−2i)=(−2i)3+2(−2i)2−(−2i)+3=8i−8+2i+3=−5+10i

步骤 4：平方差快速相乘

(−5−10i)(−5+10i)=(−5)2−(10i)2=25+100=125

解题核心逻辑

万能公式跳过了韦达定理的复杂对称式展开，把求根的二次式连乘，转化为复数代入多项式求值再相乘，是竞赛中处理根的对称式的万能技巧。

练习题1：

Let  be a polynomial with leading coefficient  whose three roots are the reciprocals of the three roots of  where  What is  in terms of  and 

练习题2

The roots of the polynomial  are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by  units. What is the volume of the new box?

答案1

万能公式回顾

首一三次多项式 \(f(x)=x^3+ax^2+bx+c\)，根为 \(p,q,r\)，则：

\(\boldsymbol{(p-t)(q-t)(r-t)=-f(t)}\)

题目条件

\(f(x)=x^3+ax^2+bx+c\)，根为 \(p,q,r\)；

\(g(x)\) 为首一多项式，根是 \(\boldsymbol{\dfrac1p,\dfrac1q,\dfrac1r}\)，求 \(g(1)\)。

步骤 1：写出 \(g(x)\) 的因式形式

\(g(x)=\left(x-\frac1p\right)\left(x-\frac1q\right)\left(x-\frac1r\right)\)

代入 \(x=1\)：

\(g(1)=\left(1-\frac1p\right)\left(1-\frac1q\right)\left(1-\frac1r\right)\)

步骤 2：通分变形

\(g(1)=\frac{(p-1)(q-1)(r-1)}{pqr}\)

步骤 3：万能公式直接求分子

令万能公式中 \(t=1\)：

\((p-1)(q-1)(r-1)=-f(1)=-(1+a+b+c)\)

步骤 4：韦达定理求分母

由韦达定理，\(f(x)\) 常数项：\(-pqr=c \implies \boldsymbol{pqr=-c}\)

步骤 5：代入化简

\(g(1)=\frac{-(1+a+b+c)}{-c}=\boldsymbol{\frac{1+a+b+c}{c}}\)

答案选 \(\boldsymbol{A}\)

答案2：

万能公式拓展（非常重要！AMC 必考）

对于非首一三次多项式 \(f(x)=Ax^3+Bx^2+Cx+D\)，根为 \(p,q,r\)，万能公式修正为：

\(\boldsymbol{(p-t)(q-t)(r-t)=-\frac{f(t)}{A}}\)

题目理解

原长方体长宽高是多项式 \(10x^3-39x^2+29x-6\) 的 3 个根 \(p,q,r\)；

每条棱加长 2，新长宽高：\(p+2,\ q+2,\ r+2\)；

新体积 = \(\boldsymbol{(p+2)(q+2)(r+2)}\)。

解题步骤

1. 把 \(p+2\) 写成 \(p-(-2)\)，令万能公式中 \(t=-2\)，首项系数 \(A=10\)：\((p+2)(q+2)(r+2)=-\frac{f(-2)}{10}\)
2. 代入 \(x=-2\) 计算多项式值：\(f(-2)=10(-2)^3-39(-2)^2+29(-2)-6=-80-156-58-6=-300\)
3. 求体积：\(V=-\frac{-300}{10}=30\)答案选 \(\boldsymbol{D}\)