Algebraic identities — (a+b)², (a−b)², (a+b)(a−b), (a+b)³, (a−b)³, (a³+b³), (a³−b³): Question bank

These questions attack the thinking, not the arithmetic. Each one targets a specific place where students slip. Cover the reveal, commit to an answer with a reason, then check. Return to the parent topic whenever a trap catches you.

True or false — justify

(a + b)^{2} = a^{2} + b^{2}

for some choice of

a, b

True — but only when

2 ab = 0

, i.e. when

a = 0

b = 0

. For all other values the middle term

2 ab

makes them differ, so as an identity it is false. (Picture: the big square of side

a + b

contains two extra

ab

rectangles beyond the

a^{2}

and

b^{2}

squares — see figure below.)

(a - b)^{2}

and

(b - a)^{2}

are always equal.

True.

b - a = - (a - b)

, and squaring kills the sign:

(- (a - b))^{2} = (a - b)^{2}

. Both expand to

a^{2} - 2 ab + b^{2}

(a + b)^{2} = (a - b)^{2}

can never happen.

False. They are equal exactly when

2 ab = - 2 ab

, i.e.

4 ab = 0

, so whenever

a = 0

b = 0

(e.g.

a = 0, b = 5

: both give

25

The identity

(a + b) (a - b) = a^{2} - b^{2}

still holds if

a

and

b

are the same number.

True. Put

a = b

: left side is

(2 a) (0) = 0

, right side is

a^{2} - a^{2} = 0

. Identities hold for every value, degenerate cases included.

a^{2} - b^{2}

can be factored using factoring even when

a^{2} < b^{2}

True. Factoring just means writing an expression as a product;

a^{2} - b^{2} = (a + b) (a - b)

regardless of size. If

a^{2} < b^{2}

the result is simply negative (e.g.

9 - 16 = (3 + 4) (3 - 4) = - 7

The sign pattern of

(a - b)^{3}

+ - + -

True.

(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}

; powers of the negative term

- b

alternate, giving alternating signs term by term.

(a + b)^{3} = a^{3} + b^{3}

is a valid identity.

False. It drops the two middle terms

3 a^{2} b + 3 a b^{2}

. Test

a = b = 1

(2)^{3} = 8

but

1 + 1 = 2

. The genuine sum-of-cubes identity is

a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})

a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})

, the trinomial

a^{2} - ab + b^{2}

can be split into two real linear factors.

False (for

b \neq = 0

). Reading it as a quadratic in

a

, a linear split would need a real value of

a

making it zero. Testing whether such a value exists uses the discriminant

Δ

(defined in the "Why" section); here

Δ = - 3 b^{2} < 0

, so no real root exists and it stays as one irreducible piece.

The coefficients in

(a + b)^{3}

come from Pascal's triangle.

True. Reading the triangle with its apex counted as row 0, the row for exponent

3

1, 3, 3, 1

— matching

a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

. This is the Binomial Theorem for

n = 3

. (See the triangle excerpt in the figure below.)

(a + b) (a - b)

and

(a - b) (a + b)

give different answers.

False. Multiplication is commutative, so both equal

a^{2} - b^{2}

. Order of the two factors never matters.

Spot the error

A student writes

(x - 4)^{2} = x^{2} - 16

. Find the flaw.

The middle term

- 2 (x) (4) = - 8 x

is missing. The misconception is treating "square" like it distributes across the minus; but

(x - 4)^{2}

means

(x - 4)

times itself, and multiplying two brackets always produces a cross-term. Correct expansion:

x^{2} - 8 x + 16

(three terms).

A student writes

(x + 3)^{2} = x^{2} + 9

and "checks" it at

x = 0

getting

9 = 9

. Why is the check useless?

x = 0

is exactly the value that kills the middle term

6 x

, so it can't detect the error. A good test uses a value where every term is alive, e.g.

x = 1

16 \neq = 10

A student factors

x^{2} + 16 = (x + 4) (x - 4)

. What went wrong?

Difference of squares needs a minus: it factors

x^{2} - 16

, not

x^{2} + 16

. Expanding

(x + 4) (x - 4)

gives

x^{2} - 16

, so it can never produce

+ 16

. A sum of squares has no real factorisation.

A student writes

a^{3} - b^{3} = (a - b) (a^{2} - ab + b^{2})

. Spot the sign slip.

The trinomial for a difference of cubes must have

+ ab

: it is

(a - b) (a^{2} + ab + b^{2})

. The

- ab

version belongs to the sum of cubes. (Expand to see: with

+ ab

the terms

- a^{2} b

and

+ a^{2} b

cancel correctly.)

A student expands

(2 y + 5)^{2} = 2 y^{2} + 20 y + 25

. What's wrong?

The first term is

(2 y)^{2} = 4 y^{2}

, not

2 y^{2}

. The whole quantity

2 y

is squared, so both the

2

and the

y

get squared.

A student says

8 y^{3} + 125

can't be a sum of cubes "because

8

isn't a cube." Correct them.

8 y^{3} = (2 y)^{3}

and

125 = 5^{3}

, so it is a sum of cubes. Look at the whole term

8 y^{3}

, not just the coefficient.

A student "verifies"

(a + b)^{2} = a^{2} + 2 ab + b^{2}

using FOIL but writes only

a^{2} + ab + b^{2}

. Which FOIL term did they lose?

FOIL is not just "write four products" — it forces you to keep the Outer (

a \cdot b

) and Inner (

b \cdot a

) as two separate products. They collapsed them into one

ab

; keeping both gives

ab + ba = 2 ab

, hence the doubled middle term.

Why questions

Why does the middle term vanish in

(a + b) (a - b)

Distributing gives four products; two of them are

- ab

and

+ ab

, exact opposites that cancel, leaving only the two squares

a^{2} - b^{2}

. This is the "handshake with cancellation" in the opening figure.

Why does

(a + b)^{3}

have four terms while

(a + b)^{2}

has three?

Each extra power of the binomial adds one more entry to the matching Pascal row; exponent

n

has

n + 1

terms. Exponent 2 → 3 terms (row

1, 2, 1

), exponent 3 → 4 terms (row

1, 3, 3, 1

Why is squaring not a linear operation, i.e. why

(a + b)^{2} \neq = a^{2} + b^{2}

A linear operation would satisfy

f (a + b) = f (a) + f (b)

, but squaring produces a cross-term

2 ab

from the product of different terms. That cross-term is precisely what breaks linearity — geometrically, the two

ab

rectangles in the area picture.

Why is the trinomial in the sum-of-cubes factorisation

a^{2} - ab + b^{2}

rather than

a^{2} + 2 ab + b^{2}

It is engineered so multiplying by

(a + b)

cancels every non-cube term. Expanding

(a + b) (a^{2} - ab + b^{2})

gives

a^{3} - a^{2} b + a b^{2} + a^{2} b - a b^{2} + b^{3}

; the pair

- a^{2} b

and

+ a^{2} b

cancel, and

+ a b^{2}

and

- a b^{2}

cancel, leaving exactly

a^{3} + b^{3}

. A

+ 2 ab

middle term would leave leftovers.

What is the "discriminant" that decides whether

a^{2} - ab + b^{2}

has real roots?

For a quadratic

A a^{2} + B a + C

(here

A = 1, B = - b, C = b^{2}

, with

b

a fixed parameter), the discriminant is

Δ = B^{2} - 4 A C

. If

Δ < 0

there is no real value of

a

making the quadratic zero, so it can't split into real linear factors. Here

Δ = (- b)^{2} - 4 (1) (b^{2}) = - 3 b^{2} < 0

(for

b \neq = 0

) ⇒ irreducible over

R

Why can completing the square rely on the identity

(a + b)^{2} = a^{2} + 2 ab + b^{2}

To turn

x^{2} + b x

into a perfect square you must supply the missing

b^{2}

-slot of the identity, namely

(b /2)^{2}

, so that

x^{2} + b x + (b /2)^{2} = (x + b /2)^{2}

— literally filling in the corner square of the area picture.

Why does knowing

(a + b) (a - b) = a^{2} - b^{2}

speed up multiplying

103 \times 97

Write them as

100 + 3

and

100 - 3

: the product is

10 0^{2} - 3^{2} = 10000 - 9 = 9991

, replacing a full multiplication with one square and one subtraction.

Why do

(a - b)^{3}

and

- (b - a)^{3}

give the same expression?

Start from

a - b = - (b - a)

. Cubing both sides:

(a - b)^{3} = (- (b - a))^{3} = (- 1)^{3} (b - a)^{3} = - (b - a)^{3}

, because

(- 1)^{3} = - 1

. An odd exponent keeps one leftover minus sign, whereas an even exponent (like squaring,

(- 1)^{2} = 1

) would erase it — that is the exact difference from

(a - b)^{2} = (b - a)^{2}

Edge cases

What is

(a + b)^{2}

when

b = 0

It collapses to

a^{2}

, since

2 ab

and

b^{2}

both vanish. The identity gracefully reduces to the trivial "square of a single term."

What does the difference-of-squares identity give when

a = b

(a + b) (a - b) = (2 a) (0) = 0

, matching

a^{2} - a^{2} = 0

. The factor

(a - b)

becomes zero, correctly signalling equal terms.

Does

a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})

hold when one term is zero, say

b = 0

Yes: right side becomes

(a) (a^{2}) = a^{3}

, equal to

a^{3} + 0

. Identities hold at every value, boundaries included.

(a + b)^{2} = a^{2} + 2 ab + b^{2}

still true if

a

b

is negative?

Yes — the identity comes from distribution alone, which knows nothing about signs. E.g.

a = 3, b = - 1

(2)^{2} = 4

and

9 - 6 + 1 = 4

. ✓

What happens to

a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})

when

a = b

The factor

(a - b) = 0

makes the whole product

0

, matching

a^{3} - a^{3} = 0

. The trinomial stays a positive

3 a^{2}

but is multiplied by zero.

Can

x^{2} - b^{2}

be a difference of squares if

b

is irrational, like

b = 5

Yes:

x^{2} - 5 = (x + 5) (x - 5)

, taking

b = 5

. The identity places no restriction on

b

being a whole number —

a, b

range over all reals.

Recall Quick self-check

The single fastest way to kill a suspected false identity is to ::: substitute small unequal numbers (like $a = 2, b = 1$ ) and compare both sides. The reason $(a + b)^{2}$ , $(a + b) (a - b)$ , and the cubes all follow from one rule is ::: the distributive property applied repeatedly. The discriminant $Δ = B^{2} - 4 A C$ of $a^{2} - ab + b^{2}$ (in $a$ ) equals ::: $- 3 b^{2}$ , which is negative, so the trinomial has no real linear factors.