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SSC CGL Maths · Algebraic Identities and Simplification

Sums and Products of Three Variables

The square of a + b + c, the sum of the squared differences, and what follows when a sum of squares is zero.

All Algebraic Identities and Simplification notes

Why this matters

11 questions in the bank, nine of them EASY. One identity links a + b + c, a² + b² + c² and ab + bc + ca, so any two of them give the third. The rest lean on one fact: a square is never negative, so squares that add up to zero must each be zero.

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

Squaring a+b+ca + b + c gives each square once and each pair product twice. Knowing any two of the sum, the sum of squares and the sum of pair products gives 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).
  • So ab+bc+ca=(a+b+c)2−(a2+b2+c2)2ab + bc + ca = \frac{(a + b + c)^2 - (a^2 + b^2 + c^2)}{2} and a2+b2+c2=(a+b+c)2−2(ab+bc+ca)a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca).
  • Recognise it in numbers: three squares plus twice each pair product is the square of the sum. Divided by the sum, it leaves the sum.
  • Fast check: with whole numbers, look for a triple. A sum of 6 with squares adding to 14 is 1, 2 and 3.

Square of a trinomial

(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=9a + b + c = 9 and a2+b2+c2=29a^2 + b^2 + c^2 = 29, find ab+bc+caab + bc + ca.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 13 Sep 2024, Shift 3 · Q54Easy

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

Simplify 1.5×1.5+2.5×2.5+3.5×3.5+2×1.5×2.5+2×2.5×3.5+2×1.5×3.51.5+2.5+3.5\frac{1.5 \times 1.5 + 2.5 \times 2.5 + 3.5 \times 3.5 + 2 \times 1.5 \times 2.5 + 2 \times 2.5 \times 3.5 + 2 \times 1.5 \times 3.5}{1.5 + 2.5 + 3.5}.

Forgetting to halve

ab+bc+caab + bc + ca is half of (a+b+c)2−(a2+b2+c2)(a + b + c)^2 - (a^2 + b^2 + c^2). Skipping the half gives double the answer, and that double is often one of the options.
Drill 4 more on the square of a + b + c

Concept 2 of 3: The sum of squared differences

(a−b)2+(b−c)2+(c−a)2(a - b)^2 + (b - c)^2 + (c - a)^2 counts every square twice and takes away every pair product twice. So it is twice the gap between a2+b2+c2a^2 + b^2 + c^2 and ab+bc+caab + bc + ca.

Definition

  • (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).
  • So a2+b2+c2−ab−bc−ca=12[(a−b)2+(b−c)2+(c−a)2]a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2}\left[(a - b)^2 + (b - c)^2 + (c - a)^2\right].
  • Numbers close together: for 71, 73 and 74 the differences are 2, 1 and 3, so a2+b2+c2−ab−bc−ca=12(4+1+9)=7a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2}(4 + 1 + 9) = 7. No big squares needed.
  • Given the sum and the sum of squares, find ab+bc+caab + bc + ca first with the previous concept, then use this one.

Sum of squared differences

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

Worked example

If a+b+c=12a + b + c = 12 and a2+b2+c2=56a^2 + b^2 + c^2 = 56, find (a−b)2+(b−c)2+(c−a)2(a - b)^2 + (b - c)^2 + (c - a)^2.
Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2023 · Tier I, 18 Jul 2023, Shift 2 · Q68Easy

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

If (a+b+c)=19(a + b + c) = 19, and (a2+b2+c2)=155(a^2 + b^2 + c^2) = 155, find the value of (a−b)2+(b−c)2+(c−a)2(a - b)^2 + (b - c)^2 + (c - a)^2.

A sum of squared differences below zero

(a−b)2+(b−c)2+(c−a)2(a - b)^2 + (b - c)^2 + (c - a)^2 is never negative for real numbers. If your working makes it negative, there is a slip in the arithmetic, or no real numbers fit the data.
Drill 2 more on the sum of squared differences

Concept 3 of 3: Squares that add to zero, and least values

The square of a real number is zero or positive. If several squares add up to zero, there is no room for any of them to be positive, so every one is zero.

Definition

  • If (a−p)2+(b−q)2+(c−r)2=0(a - p)^2 + (b - q)^2 + (c - r)^2 = 0, then a=pa = p, b=qb = q and c=rc = r.
  • Complete the squares first: move every term to one side and group. a2−4a+b2+6b+13=0a^2 - 4a + b^2 + 6b + 13 = 0 is (a−2)2+(b+3)2=0(a - 2)^2 + (b + 3)^2 = 0, so a=2a = 2 and b=−3b = -3.
  • a2+b2+c2=ab+bc+caa^2 + b^2 + c^2 = ab + bc + ca means a=b=ca = b = c. The gap between the two sides is half the sum of squared differences, which is zero only when every difference is.
  • Once the numbers are equal, put a=b=ca = b = c into the expression; the letters cancel and a number is left.
  • Least values: for positive t, t+1t≥2t + \frac{1}{t} \ge 2, because t+1t−2=(t−1t)2t + \frac{1}{t} - 2 = \left(\sqrt{t} - \frac{1}{\sqrt{t}}\right)^2. For positive x, y and z, (x+y+z)(1x+1y+1z)≥9(x + y + z)\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \ge 9, with equality when x=y=zx = y = z.

Zero sums of squares and least values

a2+b2+c2−ab−bc−ca=0⇒a=b=ct+1t≥2 (t>0)(x+y+z)(1x+1y+1z)≥9a^2 + b^2 + c^2 - ab - bc - ca = 0 \Rightarrow a = b = c \qquad t + \frac{1}{t} \ge 2 \ (t > 0) \qquad (x + y + z)\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \ge 9

Worked example

If x2+y2−6x+2y+10=0x^2 + y^2 - 6x + 2y + 10 = 0, find x2+y2x^2 + y^2.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 12 Sep 2024, Shift 3 · Q62Easy

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

If a2+b2+c2=2(a+c−1)a^2 + b^2 + c^2 = 2(a + c - 1), then the value of a3+b3+c3=a^3 + b^3 + c^3 =?

A least value needs the sign

t+1t≥2t + \frac{1}{t} \ge 2 holds only for positive t. For negative t the sum is −2-2 or less, so a least-value question must say the numbers are positive.

Equal values need the condition

a=b=ca = b = c follows from a2+b2+c2=ab+bc+caa^2 + b^2 + c^2 = ab + bc + ca, not from a+b+c=0a + b + c = 0. Put equal values in only when the given condition forces them.
Drill 2 more on squares that add to zero, and least values

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 trinomial

    (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)
  • The sum of squared differences

    Sum of squared differences

    (a−b)2+(b−c)2+(c−a)2=2(a2+b2+c2−ab−bc−ca)(a - b)^2 + (b - c)^2 + (c - a)^2 = 2(a^2 + b^2 + c^2 - ab - bc - ca)
  • Squares that add to zero, and least values

    Zero sums of squares and least values

    a2+b2+c2−ab−bc−ca=0⇒a=b=ct+1t≥2 (t>0)(x+y+z)(1x+1y+1z)≥9a^2 + b^2 + c^2 - ab - bc - ca = 0 \Rightarrow a = b = c \qquad t + \frac{1}{t} \ge 2 \ (t > 0) \qquad (x + y + z)\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \ge 9

Watch out for (4)

Test yourself on Algebraic Identities and Simplification

20 past SSC CGL questions from this chapter, timed at 12 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Algebraic Identities and Simplification chapter testDrill all 11 questions