Algebraic Identities and Simplification — SSC CGL Formulas and Traps

Squares of a Binomial

Square of a sum, square of a difference, difference of squares

Square identities

(a + b)^2 = a^2 + 2ab + b^2 \qquad (a - b)^2 = a^2 - 2ab + b^2 \qquad (a + b)^2 + (a - b)^2 = 2(a^2 + b^2) \qquad (a + b)^2 - (a - b)^2 = 4ab \qquad a^2 - b^2 = (a - b)(a + b)

Squares from the sum and product of two numbers

From a sum and a product

a^2 + b^2 = (a + b)^2 - 2ab \qquad (a - b)^2 = (a + b)^2 - 4ab \qquad a^2 - ab + b^2 = (a + b)^2 - 3ab \qquad \frac{a}{b} + \frac{b}{a} = \frac{a^2 + b^2}{ab}

Common traps

A test value that hides the difference

At

x = 1

,

x

,

x^2

and

x^3

are all 1, so

4x

and

4x^2

give the same number. When options differ only in a power, test at

x = 2

as well.

Squaring term by term

(a + b)^2

is not

a^2 + b^2

: the

2ab

is missing. At

a = b = 1

the first is 4 and the second is 2.

The last term of a squared difference

(x - 5)^2 = x^2 - 10x + 25

. The last term is

+25

, never

-25

, because it is

(-5)^2

. Only the middle term carries the minus sign.

Leaving out the coefficient in the product

With

a = 3x

and

b = 4y

,

ab = 12xy

, not

xy

. Using

xy

in

(a - b)^2 = (a + b)^2 - 4ab

gives a wrong difference.

Two square roots, one answer

(a - b)^2 = 49

gives

a - b = 7

or

-7

. The question's condition (

a > b

, natural numbers, a positive result) picks one. Without a condition, see which sign the options allow.

Cubes of a Binomial

Learn this subtopic in the notes

Cubes from a sum or difference and a product

Cube identities for two terms

(a + b)^3 = a^3 + b^3 + 3ab(a + b) \qquad (a - b)^3 = a^3 - b^3 - 3ab(a - b) \qquad a^3 + b^3 = (a + b)^3 - 3ab(a + b) \qquad a^3 - b^3 = (a - b)^3 + 3ab(a - b)

Spotting an expanded cube

Expanded cubes

(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \qquad (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3

Sum and difference of cubes as factors

Sum and difference of cubes

a^3 + b^3 = (a + b)(a^2 - ab + b^2) \qquad a^3 - b^3 = (a - b)(a^2 + ab + b^2)

Common traps

The correction carries the sum

a^3 + b^3 = (a + b)^3 - 3ab(a + b)

, not

(a + b)^3 - 3ab

. Leaving out the factor

(a + b)

is the commonest wrong option.

The sign in the difference form

a^3 - b^3 = (a - b)^3 + 3ab(a - b)

: plus, because

(a - b)^3

already took

3ab(a - b)

away. Writing minus gives a number that is too small.

Cubing a near-cube

27x^3 + 54x^2y + 36xy^2 + 8y^3

is

(3x + 2y)^3

, but change one middle coefficient and it is not. Check both middle coefficients before you cube; otherwise substitute term by term.

Pairing the wrong quadratic

a^3 + b^3

cancels with

a^2 - ab + b^2

, not with

a^2 + ab + b^2

. With the wrong pair nothing cancels and the value is not

a + b

.

Different numbers on top and below

The shortcut needs the same pair in both places. If the top cubes 0.2 and 0.3 but the bottom is built from 0.5 and 0.3, nothing cancels: work the arithmetic instead.

Reciprocal Sums x ± 1/x

Learn this subtopic in the notes

Squaring x + 1/x and x − 1/x

The squaring ladder

x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 \qquad x^2 + \frac{1}{x^2} = \left(x - \frac{1}{x}\right)^2 + 2 \qquad x^4 + \frac{1}{x^4} = \left(x^2 + \frac{1}{x^2}\right)^2 - 2

Switching between x + 1/x and x − 1/x

Switching the sign

\left(x - \frac{1}{x}\right)^2 = \left(x + \frac{1}{x}\right)^2 - 4 \qquad x^2 - \frac{1}{x^2} = \left(x + \frac{1}{x}\right)\left(x - \frac{1}{x}\right) \qquad x^4 - \frac{1}{x^4} = \left(x^2 + \frac{1}{x^2}\right)\left(x^2 - \frac{1}{x^2}\right)

Getting x + 1/x from an equation or a surd

From a quadratic to x ± 1/x

x^2 - kx + 1 = 0 \Rightarrow x + \frac{1}{x} = k \qquad x^2 - kx - 1 = 0 \Rightarrow x - \frac{1}{x} = k

Common traps

Plus 2 or minus 2

From

x + \frac{1}{x}

, subtract 2; from

x - \frac{1}{x}

, add 2.

\left(x - \frac{1}{x}\right)^2 = x^2 - 2 + \frac{1}{x^2}

, so

x^2 + \frac{1}{x^2}

is that square plus 2.

A number in front left inside

From

5y + \frac{5}{y} = 12

,

y + \frac{1}{y}

is

\frac{12}{5}

, not 12. Divide by the number in front first, then square.

Losing the sign when x is between 0 and 1

For

0 < x < 1

,

\frac{1}{x} > x

, so

x - \frac{1}{x}

and

x^2 - \frac{1}{x^2}

are negative. The positive root is printed as a wrong option.

Minus 4, not minus 2

\left(x - \frac{1}{x}\right)^2

is

\left(x + \frac{1}{x}\right)^2 - 4

. The 2 belongs to

x^2 + \frac{1}{x^2}

; the difference between the two squares is 4.

Dividing an equation whose ends do not match

x^2 - 6x + 4 = 0

divided by x gives

x + \frac{4}{x} = 6

, not

x + \frac{1}{x} = 6

. The trick needs the

x^2

coefficient and the constant to be equal.

A conjugate is the reciprocal only when the product is 1

\frac{1}{2 + \sqrt{5}}

is not

2 - \sqrt{5}

: the product is

4 - 5 = -1

, so the reciprocal is

\sqrt{5} - 2

. Work out

a^2 - b

first.

Cubes and Higher Powers of x ± 1/x

Learn this subtopic in the notes

Cubing x + 1/x and x − 1/x

Cubes of x ± 1/x

x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right) \qquad x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)^3 + 3\left(x - \frac{1}{x}\right) \qquad x^3 + \frac{k^3}{x^3} = \left(x + \frac{k}{x}\right)^3 - 3k\left(x + \frac{k}{x}\right)

Sixth and eighth powers of x ± 1/x

Sixth and eighth powers

x^6 + \frac{1}{x^6} = \left(x^3 + \frac{1}{x^3}\right)^2 - 2 \qquad x^6 + \frac{1}{x^6} = \left(x^2 + \frac{1}{x^2}\right)^3 - 3\left(x^2 + \frac{1}{x^2}\right) \qquad x^8 - \frac{1}{x^8} = \left(x^4 + \frac{1}{x^4}\right)\left(x^2 + \frac{1}{x^2}\right)\left(x^2 - \frac{1}{x^2}\right)

Fifth and seventh powers by multiplying two smaller ones

Product rule for powers of x ± 1/x

\left(x^m + \frac{1}{x^m}\right)\left(x^n + \frac{1}{x^n}\right) = x^{m+n} + \frac{1}{x^{m+n}} + x^{m-n} + \frac{1}{x^{m-n}} \qquad x^5 + \frac{1}{x^5} = \left(x^2 + \frac{1}{x^2}\right)\left(x^3 + \frac{1}{x^3}\right) - \left(x + \frac{1}{x}\right)

Common traps

Minus 3 times what

x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right)

, never

\left(x + \frac{1}{x}\right)^3 - 3

. With

x + \frac{1}{x} = 4

the answer is 52, not 61.

The sign in the difference cube

x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)^3 + 3\left(x - \frac{1}{x}\right)

: plus 3 times, not minus. The sum form and the difference form have opposite corrections.

Terms that do not multiply to 1

For

x + \frac{5}{x}

the two terms multiply to 5, so the correction is

3 \times 5 \times \left(x + \frac{5}{x}\right)

. Using 3 instead of 15 gives a printed wrong option.

The wrong correction when cubing a square

x^6 + \frac{1}{x^6} = \left(x^2 + \frac{1}{x^2}\right)^3 - 3\left(x^2 + \frac{1}{x^2}\right)

. The correction uses

x^2 + \frac{1}{x^2}

, the thing being cubed, not

x + \frac{1}{x}

.

A missing factor in an eighth-power difference

x^8 - \frac{1}{x^8}

splits into three known pieces:

x^4 + \frac{1}{x^4}

,

x^2 + \frac{1}{x^2}

and

x^2 - \frac{1}{x^2}

. Stopping after the first two leaves out the difference, which carries the surd.

Taking away the wrong rung

In

\left(x^2 + \frac{1}{x^2}\right)\left(x^3 + \frac{1}{x^3}\right)

, the extra pair is

x^1 + \frac{1}{x^1}

, because

3 - 2 = 1

. Take away the power equal to the difference of the two you multiplied, not one of them.

Sums and Products of Three Variables

Learn this subtopic in the notes

The square of a + b + c

Square of a trinomial

(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(a^2 + b^2 + c^2 - ab - bc - ca)

Squares that add to zero, and least values

Zero sums of squares and least values

a^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

Common traps

Forgetting to halve

ab + bc + ca

is half of

(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.

A sum of squared differences below zero

(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.

A least value needs the sign

t + \frac{1}{t} \ge 2

holds only for positive t. For negative t the sum is

-2

or less, so a least-value question must say the numbers are positive.

Equal values need the condition

a = b = c

follows from

a^2 + b^2 + c^2 = ab + bc + ca

, not from

a + b + c = 0

. Put equal values in only when the given condition forces them.

The Cube Identity and a + b + c = 0

Learn this subtopic in the notes

The factorisation of a³ + b³ + c³ − 3abc

The three-cube identity

a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \qquad a^3 + b^3 + c^3 - 3abc = \frac{1}{2}(a + b + c)\left[(a - b)^2 + (b - c)^2 + (c - a)^2\right] \qquad a^3 + b^3 + c^3 = s^3 - 3sq + 3abc

$s$ the sum a + b + c
$q$ the sum of pair products ab + bc + ca

When a + b + c = 0, a³ + b³ + c³ = 3abc

Three cubes when the sum is zero

a + b + c = 0 \Rightarrow a^3 + b^3 + c^3 = 3abc

Common traps

Forgetting to add 3abc back

The identity gives

a^3 + b^3 + c^3 - 3abc

. If the question asks for

a^3 + b^3 + c^3

, add

3abc

at the end.

The bracket uses the sum of squares

The second factor is

a^2 + b^2 + c^2 - (ab + bc + ca)

. Putting

(a + b + c)^2

in place of

a^2 + b^2 + c^2

adds

2(ab + bc + ca)

too much.

Equal values also satisfy the cube condition

x^3 + y^3 + z^3 = 3xyz

also holds when

x = y = z

, because then the bracket is zero. So the condition gives

x + y + z = 0

or

x = y = z

: check the options for both.

A sum that is nearly zero

In

21^3 + 9^3 - 31^3

, the bases add to

-1

, not 0, so

3abc

is not the answer. Add the bases before using the shortcut.

Rational Algebraic Expressions

Learn this subtopic in the notes

Factorise, then cancel

Factorisations for cancelling

x^2 + (p + q)x + pq = (x + p)(x + q) \qquad a^4 + a^2b^2 + b^4 = (a^2 + ab + b^2)(a^2 - ab + b^2)

Combining fractions and using a given condition

Adding two fractions

\frac{a}{p} + \frac{b}{q} = \frac{aq + bp}{pq}

Common traps

Cancelling terms, not factors

\frac{x^2 + 5}{x + 5}

does not simplify to x. Only a factor of the whole top and the whole bottom can cancel, and

x^2 + 5

has no factor

x + 5

.

A backwards factor changes the sign

\frac{2x - 7}{7 - 2x} = -1

, not 1, because

7 - 2x = -(2x - 7)

.

The minus sign before a fraction

In

\frac{a}{2} - \frac{b - 3}{4}

, the minus applies to the whole top of the second fraction: the result is

\frac{2a - b + 3}{4}

, not

\frac{2a - b - 3}{4}

.

Test values that break the condition

If the question says

a + b + c = 0

, testing with

a = b = c = 1

is not allowed, because those values add to 3. Use values that fit, such as 1, 2 and −3.

Algebraic Identities and Simplification formulas

Squares of a Binomial

Square of a sum, square of a difference, difference of squares

Squares from the sum and product of two numbers

Common traps

Cubes of a Binomial

Cubes from a sum or difference and a product

Spotting an expanded cube

Sum and difference of cubes as factors

Common traps

Reciprocal Sums x ± 1/x

Squaring x + 1/x and x − 1/x

Switching between x + 1/x and x − 1/x

Getting x + 1/x from an equation or a surd

Common traps

Cubes and Higher Powers of x ± 1/x

Cubing x + 1/x and x − 1/x

Sixth and eighth powers of x ± 1/x

Fifth and seventh powers by multiplying two smaller ones

Common traps

Sums and Products of Three Variables

The square of a + b + c

The sum of squared differences

Squares that add to zero, and least values

Common traps

The Cube Identity and a + b + c = 0

The factorisation of a³ + b³ + c³ − 3abc

When a + b + c = 0, a³ + b³ + c³ = 3abc

Common traps

Rational Algebraic Expressions

Factorise, then cancel

Combining fractions and using a given condition

Common traps

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