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

Here is the toolbox we will keep reaching for. Every symbol below was earned in the parent note; this is just the shelf we pick from.

Level 1 — Recognition

Goal: given an expression, name which identity it is a case of. No heavy algebra yet — just matching.

Exercise 1.1

Which identity does $(m + 7)^{2}$ match, and what are $a$ and $b$ ?

Recall Solution 1.1

The expression is a binomial being squared, with a plus sign inside. That is the square of a sum, Identity 1: $(a + b)^{2} = a^{2} + 2 ab + b^{2} .$ Matching term by term: the first slot $a$ is filled by $m$ , the second slot $b$ by $7$ . Answer: $a = m$ , $b = 7$ .

Exercise 1.2

The expression $p^{2} - 49$ is a case of which identity? Identify $a$ and $b$ .

Recall Solution 1.2

Two things separated by a minus, and each piece is a perfect square: $p^{2} = (p)^{2}$ and $49 = 7^{2}$ . That signature — square minus square — is the difference of squares, Identity 3 read backwards: $a^{2} - b^{2} = (a + b) (a - b) .$ Answer: $a = p$ , $b = 7$ (since $49 = 7^{2}$ ).

Exercise 1.3

Is $8 k^{3} + 1$ a sum of cubes? If so, what are $a$ and $b$ ?

Recall Solution 1.3

Ask: is each piece a perfect cube? $8 k^{3} = (2 k)^{3}$ because $2^{3} = 8$ and $(k)^{3} = k^{3}$ . And $1 = 1^{3}$ . Two cubes joined by a plus — yes, this is the sum of cubes, Identity 6. Answer: $a = 2 k$ , $b = 1$ .

Level 2 — Application

Goal: actually expand or factor by substituting into the identity and simplifying.

Exercise 2.1

Expand $(3 x + 4)^{2}$ .

Recall Solution 2.1

Which identity and why: it's a sum being squared → Identity 1, $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ . We choose this over multiplying out longhand because the identity gives the three terms directly. Take $a = 3 x$ , $b = 4$ . $(3 x + 4)^{2} = (3 x)^{2} + 2 (3 x) (4) + 4^{2}$ Now simplify each piece. $(3 x)^{2} = 3^{2} x^{2} = 9 x^{2}$ (the exponent hits both the $3$ and the $x$ ). The middle term $2 (3 x) (4) = 24 x$ . The last, $4^{2} = 16$ . $(3 x + 4)^{2} = 9 x^{2} + 24 x + 16.$

Exercise 2.2

Expand $(5 - 2 n)^{2}$ .

Recall Solution 2.2

A difference squared → Identity 2, $(a - b)^{2} = a^{2} - 2 ab + b^{2}$ , so the middle term is negative. Take $a = 5$ , $b = 2 n$ . $(5 - 2 n)^{2} = 5^{2} - 2 (5) (2 n) + (2 n)^{2}$ $= 25 - 20 n + 4 n^{2}$ . Notice the last term is $+ 4 n^{2}$ , positive, because $b^{2} = (2 n)^{2}$ is a square and squares are never negative. $(5 - 2 n)^{2} = 25 - 20 n + 4 n^{2} .$

Exercise 2.3

Factor $x^{3} - 64$ .

Recall Solution 2.3

Two terms, minus sign, each a cube: $x^{3} = (x)^{3}$ and $64 = 4^{3}$ (because $4 \times 4 \times 4 = 64$ ). → Difference of cubes, Identity 7: $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2}) .$ With $a = x$ , $b = 4$ : $x^{3} - 64 = (x - 4) (x^{2} + 4 x + 16) .$ (In the trinomial the middle sign is plus for a difference of cubes — that's the part everyone flips.)

Exercise 2.4

Expand $(2 t + 1)^{3}$ .

Recall Solution 2.4

Sum being cubed → Identity 4, $(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ . The coefficients $1, 3, 3, 1$ come from Pascal's Triangle row 3. Take $a = 2 t$ , $b = 1$ . $(2 t + 1)^{3} = (2 t)^{3} + 3 (2 t)^{2} (1) + 3 (2 t) (1)^{2} + 1^{3}$ Simplify: $(2 t)^{3} = 8 t^{3}$ ; $3 (2 t)^{2} (1) = 3 (4 t^{2}) = 12 t^{2}$ ; $3 (2 t) (1)^{2} = 6 t$ ; $1^{3} = 1$ . $(2 t + 1)^{3} = 8 t^{3} + 12 t^{2} + 6 t + 1.$

Level 3 — Analysis

Goal: reverse-engineer, combine two identities, or use them for clever arithmetic.

Exercise 3.1

Factor completely: $y^{4} - 81$ .

Recall Solution 3.1

$y^{4} = (y^{2})^{2}$ and $81 = 9^{2}$ , so this is a difference of squares with $a = y^{2}$ , $b = 9$ : $y^{4} - 81 = (y^{2} + 9) (y^{2} - 9) .$ Don't stop — the checklist says look at each factor again. The second factor $y^{2} - 9$ is itself a difference of squares ( $a = y$ , $b = 3$ ): $y^{2} - 9 = (y + 3) (y - 3) .$ The first factor $y^{2} + 9$ is a sum of squares — not one of our seven identities, and it cannot be factored with real numbers, so we leave it. $y^{4} - 81 = (y^{2} + 9) (y + 3) (y - 3) .$

Exercise 3.2

Use the difference-of-squares identity to compute $103 \times 97$ in your head.

Recall Solution 3.2

Why this tool: $103$ and $97$ sit symmetrically around $100$ — one is $100 + 3$ , the other $100 - 3$ . That is exactly the shape $(a + b) (a - b)$ , which collapses to $a^{2} - b^{2}$ . See the number line below.

Figure — Algebraic identities — (a+b)², (a−b)², (a+b)(a−b), (a+b)³, (a−b)³, (a³+b³), (a³−b³)

Take $a = 100$ , $b = 3$ : $103 \times 97 = (100 + 3) (100 - 3) = 10 0^{2} - 3^{2} = 10000 - 9 = 9991.$ Answer: $9991$ .

Exercise 3.3

Given $a + b = 7$ and $ab = 10$ , find $a^{2} + b^{2}$ without solving for $a$ and $b$ .

Recall Solution 3.3

The idea: Identity 1 links these exact quantities. Start from $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ and rearrange to isolate $a^{2} + b^{2}$ : $a^{2} + b^{2} = (a + b)^{2} - 2 ab .$ This works because $(a + b)^{2}$ already contains $a^{2} + b^{2}$ plus the extra $2 ab$ ; subtracting $2 ab$ peels off the extra. Substitute the given values: $a^{2} + b^{2} = 7^{2} - 2 (10) = 49 - 20 = 29.$ Answer: $29$ .

Level 4 — Synthesis

Goal: chain several identities and manage signs across a longer argument.

Exercise 4.1

Simplify $(x + 2)^{2} - (x - 2)^{2}$ .

Recall Solution 4.1

Two routes; both must agree. Route A — expand each: $(x + 2)^{2} = x^{2} + 4 x + 4, (x - 2)^{2} = x^{2} - 4 x + 4.$ Subtract, being careful to flip every sign of the second bracket: $(x^{2} + 4 x + 4) - (x^{2} - 4 x + 4) = x^{2} + 4 x + 4 - x^{2} + 4 x - 4 = 8 x .$ Route B — spot difference of squares: the whole thing is $P^{2} - Q^{2}$ with $P = x + 2$ , $Q = x - 2$ , so it equals $(P + Q) (P - Q)$ : $[(x + 2) + (x - 2)] [(x + 2) - (x - 2)] = (2 x) (4) = 8 x .$ Both give $8 x$ . ✓

Exercise 4.2

Factor $x^{6} - y^{6}$ completely. (Hint: it's both a difference of squares and a difference of cubes.)

Recall Solution 4.2

$x^{6} = (x^{3})^{2} = (x^{2})^{3}$ , so we can slice it two ways. Take the difference-of-squares view first, because it gives two smaller cubic-shaped factors: $x^{6} - y^{6} = (x^{3})^{2} - (y^{3})^{2} = (x^{3} + y^{3}) (x^{3} - y^{3}) .$ Now each factor is a cube pattern. Sum of cubes (Identity 6) with $a = x, b = y$ : $x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2}) .$ Difference of cubes (Identity 7): $x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2}) .$ Putting it together: $x^{6} - y^{6} = (x + y) (x^{2} - x y + y^{2}) (x - y) (x^{2} + x y + y^{2}) .$ The two trinomials cannot be factored further with real numbers, so we stop.

Exercise 4.3

If $a - b = 3$ and $ab = 4$ , find $a^{3} - b^{3}$ .

Recall Solution 4.3

Plan: Identity 7 says $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$ . We already have $a - b$ and $ab$ ; we just need $a^{2} + b^{2}$ , which the square-of-a-difference identity supplies: $a^{2} + b^{2} = (a - b)^{2} + 2 ab = 3^{2} + 2 (4) = 9 + 8 = 17.$ (Here we rearranged $(a - b)^{2} = a^{2} - 2 ab + b^{2}$ to get $a^{2} + b^{2} = (a - b)^{2} + 2 ab$ .) Then $a^{2} + ab + b^{2} = (a^{2} + b^{2}) + ab = 17 + 4 = 21$ . Finally: $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2}) = 3 \times 21 = 63.$ Answer: $63$ .

Level 5 — Mastery

Goal: prove, generalise, and connect the identities to bigger ideas.

Exercise 5.1

Prove the identity $a^{2} + b^{2} + c^{2} - ab - b c - c a = \frac{1}{2} [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}]$ , and use it to show the left side is never negative for real $a, b, c$ .

Recall Solution 5.1

Start from the messier-looking right side and expand using Identity 2 three times: $(a - b)^{2} = a^{2} - 2 ab + b^{2},$ $(b - c)^{2} = b^{2} - 2 b c + c^{2},$ $(c - a)^{2} = c^{2} - 2 c a + a^{2} .$ Add all three: $= 2 a^{2} + 2 b^{2} + 2 c^{2} - 2 ab - 2 b c - 2 c a .$ Multiply by $\frac{1}{2}$ : $\frac{1}{2} [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}] = a^{2} + b^{2} + c^{2} - ab - b c - c a .$ That is exactly the left side. ✓ Why it's never negative: each square $(a - b)^{2}, (b - c)^{2}, (c - a)^{2}$ is $\geq 0$ (a real number squared can't be negative), their sum is $\geq 0$ , and half of a non-negative number is still non-negative. So the whole expression $\geq 0$ , hitting $0$ only when $a = b = c$ .

Exercise 5.2

Show that $n^{3} - n$ is divisible by $6$ for every integer $n$ .

Recall Solution 5.2

Factor first, because factors reveal divisibility. Pull out $n$ , then use difference of squares on $n^{2} - 1$ : $n^{3} - n = n (n^{2} - 1) = n (n - 1) (n + 1) = (n - 1) n (n + 1) .$ This is a product of three consecutive integers.

Among any three consecutive integers, one is divisible by $3$ → the product is divisible by $3$ .
Among any two consecutive integers there is an even one, so among three there is at least one even → the product is divisible by $2$ . Divisible by both $2$ and $3$ , and since $2$ and $3$ share no common factor, the product is divisible by $2 \times 3 = 6$ . ∎ (Quick check: $n = 4$ gives $64 - 4 = 60 = 6 \times 10$ . ✓)

Exercise 5.3

The identity $a^{2} + 2 ab + b^{2} = (a + b)^{2}$ read backwards powers a whole technique: Completing the Square. Rewrite $x^{2} + 10 x + 7$ in the form $(x + p)^{2} + q$ and state $p$ and $q$ .

Recall Solution 5.3

Why this works: we want $x^{2} + 10 x$ to be the start of a perfect square $x^{2} + 2 ab + b^{2}$ . Compare $2 ab$ with $10 x$ taking $a = x$ : then $2 b = 10$ , so $b = 5$ . The completed square would be $(x + 5)^{2} = x^{2} + 10 x + 25$ . But our expression only has $+ 7$ , not $+ 25$ . So we add and subtract the missing amount: $x^{2} + 10 x + 7 = (x + 5)^{2} (x^{2} + 10 x + 25) - 25 + 7 = (x + 5)^{2} - 18.$ Answer: $p = 5$ , $q = - 18$ . This is the exact move that solves general Quadratic Equations.