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SPI Mathematics · Algebraic Identities and Simplification

Finding Values with Identities

An identity turns a given sum and product, or a condition such as a + b + c = 0, straight into the value asked for, without finding the letters themselves.

All Algebraic Identities and Simplification notes

Why this matters

Five questions in the bank, two EASY and three MODERATE, and both of this chapter's questions from the 2022 paper are here. None needs the values of the letters: each is one identity, or a chain of brackets that cancels, and the work is choosing which.

Concept 1 of 3: Squares and cubes from the sum and the product

If you know x+yx + y and xyxy, you know x2+y2x^2 + y^2 and x3+y3x^3 + y^3 too, because each can be rewritten using only the sum and the product. There is no need to find x and y.

Definition

  • x2+y2=(x+y)2−2xy=(x−y)2+2xyx^2 + y^2 = (x + y)^2 - 2xy = (x - y)^2 + 2xy.
  • x3+y3=(x+y)3−3xy(x+y)x^3 + y^3 = (x + y)^3 - 3xy(x + y) and x3−y3=(x−y)3+3xy(x−y)x^3 - y^3 = (x - y)^3 + 3xy(x - y).
  • With y=1xy = \frac{1}{x} the product is 1: x2+1x2=(x+1x)2−2x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2.

Sum and product identities

x2+y2=(x+y)2−2xyx3+y3=(x+y)3−3xy(x+y)x3−y3=(x−y)3+3xy(x−y)x^2 + y^2 = (x + y)^2 - 2xy \qquad x^3 + y^3 = (x + y)^3 - 3xy(x + y) \qquad x^3 - y^3 = (x - y)^3 + 3xy(x - y)

Worked example

If a+b=7a + b = 7 and ab=10ab = 10, find a2+b2a^2 + b^2 and a3+b3a^3 + b^3.
Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionEasy

Example 1 · Algebraic Identities and Simplification · Finding Values with Identities

If a + b = 8 and ab = 6 then the value of a3+b3a^3 + b^3 =

Squaring the sum and stopping

(x+y)2(x + y)^2 is x2+y2+2xyx^2 + y^2 + 2xy, not x2+y2x^2 + y^2. Subtract 2xy2xy: with x+y=6x + y = 6 and xy=5xy = 5, x2+y2=36−10=26x^2 + y^2 = 36 - 10 = 26.

Leaving out the bracket in the cube identity

x3+y3=(x+y)3−3xy(x+y)x^3 + y^3 = (x + y)^3 - 3xy(x + y). The term taken away is 3xy3xy times the sum, not just 3xy3xy.
Drill 1 more on squares and cubes from the sum and the product

Concept 2 of 3: Three-term identities and the condition a + b + c = 0

Squaring a+b+ca + b + c gives every square once and every pair product twice. And the cube identity has a+b+ca + b + c as a factor, so when the three add to 0, the sum of the cubes is simply 3abc3abc.

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).
  • a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca).
  • So if a+b+c=0a + b + c = 0: a3+b3+c3=3abca^3 + b^3 + c^3 = 3abc and a2+b2+c2=−2(ab+bc+ca)a^2 + b^2 + c^2 = -2(ab + bc + ca).
  • Take each term with its sign. A condition such as x=2y+6x = 2y + 6 is x+(−2y)+(−6)=0x + (-2y) + (-6) = 0.

Three-term identities

(a+b+c)2=a2+b2+c2+2(ab+bc+ca)a+b+c=0⇒a3+b3+c3=3abc(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \qquad a + b + c = 0 \Rightarrow a^3 + b^3 + c^3 = 3abc

Worked example

Find the value of 153−83−7315^3 - 8^3 - 7^3 without cubing.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SPI · 2022 · SPI Entrance Examination 2022 · Q66Moderate

Example 2 · Algebraic Identities and Simplification · Finding Values with Identities

If a+b+c=0a + b + c = 0, then a2+b2+c2bc+ca+ab=?\frac{a^2 + b^2 + c^2}{bc + ca + ab} = ?

Forgetting the 2 in the square of three terms

(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). Without the 2, a2+b2+c2a^2 + b^2 + c^2 comes out wrong by ab+bc+caab + bc + ca.

Dropping the signs of the three terms

In p3−27q3−1p^3 - 27q^3 - 1 the three numbers being cubed are p, −3q-3q and −1-1, not p, 3q and 1. Their sum and their product must use those signs.
Drill 1 more on three-term identities and the condition a + b + c = 0

Concept 3 of 3: Simplifying a chain of brackets that cancel

A bracket like 1+1n1 + \frac{1}{n} is the single fraction n+1n\frac{n + 1}{n}. Write a whole chain of such brackets as fractions, and the top of each one cancels the bottom of the next, leaving almost nothing.

Definition

  • 1+1n=n+1n1 + \frac{1}{n} = \frac{n + 1}{n} and 1−1n=n−1n1 - \frac{1}{n} = \frac{n - 1}{n}.
  • In a chain of these, each numerator cancels the next denominator. Only one number from each end survives: 54×65×76=74\frac{5}{4} \times \frac{6}{5} \times \frac{7}{6} = \frac{7}{4}.
  • Do the cancelling first and any outside step (multiply by p, subtract 1) last.

One bracket as one fraction

1+1n=n+1n1−1n=n−1n1 + \frac{1}{n} = \frac{n + 1}{n} \qquad 1 - \frac{1}{n} = \frac{n - 1}{n}

Worked example

Simplify (1−12)(1−13)(1−14)⋯(1−110)\left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right) \cdots \left(1 - \frac{1}{10}\right).
Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SPI · 2022 · SPI Entrance Examination 2022 · Q35Moderate

Example 3 · Algebraic Identities and Simplification · Finding Values with Identities

p[(1+1p)(1+1p+1)(1+1p+2)−1=?p\Big[\left(1+\frac{1}{p}\right)\left(1+\frac{1}{p+1}\right)\left(1+\frac{1}{p+2}\right)-1=?

Multiplying the brackets out

Expanding (1+13)(1+14)\left(1 + \frac{1}{3}\right)\left(1 + \frac{1}{4}\right) term by term is slow and error-prone. As fractions it is 43×54=53\frac{4}{3} \times \frac{5}{4} = \frac{5}{3} at once.

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)

  • Squares and cubes from the sum and the product

    Sum and product identities

    x2+y2=(x+y)2−2xyx3+y3=(x+y)3−3xy(x+y)x3−y3=(x−y)3+3xy(x−y)x^2 + y^2 = (x + y)^2 - 2xy \qquad x^3 + y^3 = (x + y)^3 - 3xy(x + y) \qquad x^3 - y^3 = (x - y)^3 + 3xy(x - y)
  • Three-term identities and the condition a + b + c = 0

    Three-term identities

    (a+b+c)2=a2+b2+c2+2(ab+bc+ca)a+b+c=0⇒a3+b3+c3=3abc(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \qquad a + b + c = 0 \Rightarrow a^3 + b^3 + c^3 = 3abc
  • Simplifying a chain of brackets that cancel

    One bracket as one fraction

    1+1n=n+1n1−1n=n−1n1 + \frac{1}{n} = \frac{n + 1}{n} \qquad 1 - \frac{1}{n} = \frac{n - 1}{n}

Watch out for (5)

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