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CDS Mathematics · Algebraic Identities and Simplification

Sums and Products of Three Variables

The square of a + b + c links the sum of squares to the sum of pairwise products, and expanding a product of brackets is read off from those same sums.

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Why this matters

Eleven PYQs, mostly MODERATE. Three pieces carry the page: (a + b + c)² = Σa² + 2Σab, the expansion of (x − a)(x − b)(x − c), and the substitution x = s − a when 2s = a + b + c. The 'sum of products two at a time' of a list of numbers is the first identity in disguise.

Concept 1 of 3: The square of a + b + c

Squaring a+b+ca + b + c produces each square once and each product of two different letters twice. So any two of ∑a\sum a, ∑a2\sum a^2, ∑ab\sum ab fix the third.

Definition

  • (a+b+c)2=a2+b2+c2+2(ab+bc+ca)(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca).
  • For any list of numbers, the sum of all products two at a time is (sum)2−sum of squares2\dfrac{(\text{sum})^2 - \text{sum of squares}}{2}.
  • (a−b)2+(b−c)2+(c−a)2=2(a2+b2+c2)−2(ab+bc+ca)(a - b)^2 + (b - c)^2 + (c - a)^2 = 2(a^2 + b^2 + c^2) - 2(ab + bc + ca).

Square of a sum of three

(a+b+c)2=a2+b2+c2+2(ab+bc+ca)(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)

Worked example

If a+b+c=6a + b + c = 6 and ab+bc+ca=11ab + bc + ca = 11, find a2+b2+c2a^2 + b^2 + c^2.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2025 · CDS (II) 2025 — Elementary Mathematics · Q27Moderate

Example 1 · Algebraic Identities and Simplification · Sums and Products of Three Variables

If (a−b)2+(b−c)2+(c−a)2=6(a-b)^2+(b-c)^2+(c-a)^2=6 and a2+b2+c2=29a^2+b^2+c^2=29, then what is (a+b+c)(a+b+c) equal to?

A square root gives two signs

From (a+b+c)2=64(a + b + c)^2 = 64, a+b+c=±8a + b + c = \pm 8. Keep both unless the question says the numbers are positive.
Drill 4 more on the square of a + b + c

Concept 2 of 3: Expanding a product of brackets

Multiplying out (x−a)(x−b)(x−c)(x - a)(x - b)(x - c) chooses one term from each bracket in every possible way. Collected by powers of xx, the coefficients are the sum, the sum of pairs and the product, with alternating signs.

Definition

  • (x−a)(x−b)(x−c)=x3−(a+b+c)x2+(ab+bc+ca)x−abc(x - a)(x - b)(x - c) = x^3 - (a + b + c)x^2 + (ab + bc + ca)x - abc.
  • Put x=1x = 1: (1−a)(1−b)(1−c)=1−∑a+∑ab−abc(1 - a)(1 - b)(1 - c) = 1 - \sum a + \sum ab - abc.
  • (a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)(a + b + c)^3 = a^3 + b^3 + c^3 + 3(a + b)(b + c)(c + a).
  • A product of brackets of distinct symbols has as many terms as the product of the bracket lengths.

Cubic from its roots

(x−a)(x−b)(x−c)=x3−∑a x2+∑ab x−abc(x - a)(x - b)(x - c) = x^3 - \textstyle\sum a\, x^2 + \sum ab\, x - abc

Worked example

If a+b+c=6a + b + c = 6, ab+bc+ca=11ab + bc + ca = 11 and abc=6abc = 6, find (2−a)(2−b)(2−c)(2 - a)(2 - b)(2 - c).
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2021 · CDS (II) 2021 — Elementary Mathematics · Q12Moderate

Example 2 · Algebraic Identities and Simplification · Sums and Products of Three Variables

If α+β+γ=αβ+βγ+γα\alpha+\beta+\gamma=\alpha\beta+\beta\gamma+\gamma\alpha, then what is (1−α)(1−β)(1−γ)(1-\alpha)(1-\beta)(1-\gamma) equal to ?

The signs alternate

In (x−a)(x−b)(x−c)(x - a)(x - b)(x - c) the coefficients go +,−,+,−+, -, +, -: the xx term is +∑ab+\sum ab and the constant is −abc-abc. With (x+a)(x+b)(x+c)(x + a)(x + b)(x + c) every sign is ++.
Drill 3 more on expanding a product of brackets

Concept 3 of 3: When 2s = a + b + c

The quantities s−as - a, s−bs - b, s−cs - c are simpler than they look: any two of them add to the third side, and all three add to ss. Renaming them turns a messy expression into a short one.

Definition

If 2s=a+b+c2s = a + b + c, write x=s−ax = s - a, y=s−by = s - b, z=s−cz = s - c. Then:

  • x+y=cx + y = c, y+z=ay + z = a, z+x=bz + x = b;
  • x+y+z=sx + y + z = s;
  • (s−a)(s−b)+(s−b)(s−c)+(s−c)(s−a)=3s2−2s(a+b+c)+∑ab=∑ab−s2(s - a)(s - b) + (s - b)(s - c) + (s - c)(s - a) = 3s^2 - 2s(a + b + c) + \sum ab = \sum ab - s^2.

Semi-perimeter pieces

(s−a)+(s−b)+(s−c)=s(s - a) + (s - b) + (s - c) = s

Worked example

If 2s=a+b+c2s = a + b + c, write s2+(s−a)2+(s−b)2+(s−c)2s^2 + (s - a)^2 + (s - b)^2 + (s - c)^2 in terms of aa, bb, cc.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2023 · CDS (II) 2023 — Elementary Mathematics · Q80Moderate

Example 3 · Algebraic Identities and Simplification · Sums and Products of Three Variables

If 2s=a+b+c2s = a + b + c, then what is s2+(s−a)(s−b)+(s−b)(s−c)+(s−c)(s−a)s^2 + (s - a)(s - b) + (s - b)(s - c) + (s - c)(s - a) equal to ?

s is half the perimeter

2s=a+b+c2s = a + b + c, so s−a=b+c−a2s - a = \dfrac{b + c - a}{2}, not b+c−ab + c - a. Substituting s=a+b+cs = a + b + c doubles every term.
Drill 1 more on when 2s = a + b + c

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • The square of a + b + c

    Square of a sum of three

    (a+b+c)2=a2+b2+c2+2(ab+bc+ca)(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
  • Expanding a product of brackets

    Cubic from its roots

    (x−a)(x−b)(x−c)=x3−∑a x2+∑ab x−abc(x - a)(x - b)(x - c) = x^3 - \textstyle\sum a\, x^2 + \sum ab\, x - abc
  • When 2s = a + b + c

    Semi-perimeter pieces

    (s−a)+(s−b)+(s−c)=s(s - a) + (s - b) + (s - c) = s

Watch out for (3)

Test yourself on Algebraic Identities and Simplification

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