Understanding Algebraic Formulas: A Student Guide

Algebraic formulae in elementary mathematics describe specific relationships among algebraic quantities such as sums, differences, products, or powers, allowing for rapid calculation and manipulation of expressions. These formulae underpin core manipulations in secondary-level algebra and are foundational to advanced problem-solving.

Algebraic Expansion of the Square of a Binomial

The expression for the square of a binomial, $(a + b)^2$, can be deduced using the distributive property of multiplication over addition as follows.

Begin with $(a + b)^2$. By definition,

$(a + b)^2 = (a + b) \times (a + b)$

Apply the distributive property to expand the product:

$= a \times (a + b) + b \times (a + b)$

Expand both terms:

$= (a \times a + a \times b) + (b \times a + b \times b)$

Arrange and group like terms:

$= a^2 + ab + ba + b^2$

Since $ab = ba$, combine:

$= a^2 + 2ab + b^2$

Result: $(a + b)^2 = a^2 + 2ab + b^2$

For the square of a difference, $(a - b)^2$:

$(a - b)^2 = (a - b)\times(a - b)$

Apply the distributive property:

$= a \times (a - b) - b \times (a - b)$

Expand both terms:

$= (a \times a - a \times b) - (b \times a - b \times b)$

$= a^2 - ab - ba + b^2$

Combine $-ab$ and $-ba = -2ab$:

$= a^2 - 2ab + b^2$

Result: $(a - b)^2 = a^2 - 2ab + b^2$

An in-depth review of binomial operations is provided in Understanding Algebra Concepts.

Expansion and Deduction of the Cube of a Binomial

The expansion for the cube of a binomial, $(a + b)^3$, follows from the definition of exponentiation and distributive multiplication.

Write $ (a + b)^3 = (a + b) \times (a + b) \times (a + b) $

First, expand $(a + b) \times (a + b)$ as previously shown:

$= (a^2 + 2ab + b^2) \times (a + b)$

Distribute each term in $(a^2 + 2ab + b^2)$ over $(a + b)$:

$= a^2 \times (a + b) + 2ab \times (a + b) + b^2 \times (a + b)$

Expand each term:

$= (a^2 \times a + a^2 \times b) + (2ab \times a + 2ab \times b) + (b^2 \times a + b^2 \times b)$

$= (a^3 + a^2b) + (2a^2b + 2ab^2) + (ab^2 + b^3)$

Combine like terms:

$a^3$ appears once.

$a^2b$ and $2a^2b$ sum to $3a^2b$.

$ab^2$ and $2ab^2$ sum to $3ab^2$.

$b^3$ appears once.

$= a^3 + 3a^2b + 3ab^2 + b^3$

Result: $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$

For the cube of a difference, $(a - b)^3$:

Apply the sign alternation explicitly:

$(a - b)^3 = (a - b) \times (a - b) \times (a - b)$

$(a - b) \times (a - b) = a^2 - 2ab + b^2$ (as derived above)

Expand:

$= (a^2 - 2ab + b^2)\times(a - b)$

Distribute each term:

$= a^2 \times (a - b) - 2ab \times (a - b) + b^2 \times (a - b)$

Expand individually:

$= (a^2a - a^2b) - (2aba - 2abb) + (b^2a - b^2b)$

$= (a^3 - a^2b) - (2a^2b - 2ab^2) + (ab^2 - b^3)$

$= a^3 - a^2b - 2a^2b + 2ab^2 + ab^2 - b^3$

Combine like terms:

$a^3$ appears once.

$-a^2b - 2a^2b = -3a^2b$

$2ab^2 + ab^2 = 3ab^2$

$-b^3$ remains.

$= a^3 - 3a^2b + 3ab^2 - b^3$

Result: $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$

Algebraic Expansion of a Binomial Product

The product of the sum and difference of two terms is characterized by the following identity:

$(a + b)(a - b)$

Expand using distributivity:

$= a(a - b) + b(a - b)$

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

$= a^2 - ab + ab - b^2$

$= a^2 - b^2$ (since $-ab + ab = 0$)

Result: $(a + b)(a - b) = a^2 - b^2$

Explicit Expansion of the Square of a Trinomial

For a trinomial $(a + b + c)^2$, the expansion utilizes the commutative and distributive properties of multiplication:

$(a + b + c)^2 = (a + b + c)(a + b + c)$

Expand the first bracket term-wise:

$= a(a + b + c) + b(a + b + c) + c(a + b + c)$

Expand each:

$= (a^2 + ab + ac) + (ba + b^2 + bc) + (ca + cb + c^2)$

Group like terms:

$a^2$, $b^2$, $c^2$ each appear once.

$ab$ and $ba$ combine to $2ab$.

$ac$ and $ca$ combine to $2ac$.

$bc$ and $cb$ combine to $2bc$.

$= a^2 + b^2 + c^2 + 2ab + 2bc + 2ac$

Result: $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac$

For additional resources, refer to Key Algebraic Formulas.

Algebraic Expansion of the Cube of a Trinomial

To determine $(a + b + c)^3$, express as a product:

$(a + b + c)^3 = (a + b + c)(a + b + c)(a + b + c)$

First, expand $(a + b + c)(a + b + c)$ as derived for the square:

$= (a^2 + b^2 + c^2 + 2ab + 2bc + 2ca)\times (a + b + c)$

Distribute each term in the bracket over $(a + b + c)$:

$= a^2(a + b + c) + b^2(a + b + c) + c^2(a + b + c) + 2ab(a + b + c) + 2bc(a + b + c) + 2ca(a + b + c)$

Expand every term:

$a^2(a) = a^3$, $a^2(b) = a^2b$, $a^2(c) = a^2c$

$b^2(a) = ab^2$, $b^2(b) = b^3$, $b^2(c) = b^2c$

$c^2(a) = ac^2$, $c^2(b) = bc^2$, $c^2(c) = c^3$

$2ab(a) = 2a^2b$, $2ab(b) = 2ab^2$, $2ab(c) = 2abc$

$2bc(a) = 2abc$, $2bc(b) = 2b^2c$, $2bc(c) = 2bc^2$

$2ca(a) = 2a^2c$, $2ca(b) = 2abc$, $2ca(c) = 2ac^2$

Now, collect like terms:

$a^3$, $b^3$, $c^3$ each appear once.

For $a^2b$: $a^2b$ from $a^2(b)$ and $2a^2b$ from $2ab(a)$, sum to $3a^2b$.

For $a^2c$: $a^2c$ from $a^2(c)$ and $2a^2c$ from $2ca(a)$, sum to $3a^2c$.

For $ab^2$: $ab^2$ from $b^2(a)$ and $2ab^2$ from $2ab(b)$, sum to $3ab^2$.

For $ac^2$: $ac^2$ from $c^2(a)$ and $2ac^2$ from $2ca(c)$, sum to $3ac^2$.

For $b^2c$: $b^2c$ from $b^2(c)$ and $2b^2c$ from $2bc(b)$, sum to $3b^2c$.

For $bc^2$: $bc^2$ from $c^2(b)$ and $2bc^2$ from $2bc(c)$, sum to $3bc^2$.

For $abc$: $2abc$ from $2ab(c)$, $2abc$ from $2bc(a)$, $2abc$ from $2ca(b)$, sum to $6abc$.

Thus,

$= a^3 + b^3 + c^3 + 3(a^2b + a^2c + ab^2 + ac^2 + b^2c + bc^2) + 6abc$

Result: $(a + b + c)^3 = a^3 + b^3 + c^3 + 3(a^2b + a^2c + ab^2 + ac^2 + b^2c + bc^2) + 6abc$

Further trinomial expansions are discussed in Exploring Algebra of Functions.

Factorisation of the Difference and Sum of Cubes

The difference of cubes can be factorized as follows:

Start with $a^3 - b^3$.

Observe that

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Verification by expansion:

Multiply $(a - b)$ with $(a^2 + ab + b^2)$:

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

$= (a^3 + a^2b + ab^2) - (a^2b + ab^2 + b^3)$

$= a^3 + a^2b + ab^2 - a^2b - ab^2 - b^3$

$= a^3 - b^3$ (since $+a^2b - a^2b = 0$ and $+ab^2 - ab^2 = 0$)

Result: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Similarly,

$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

Verification is as follows:

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

$= a^3 - a^2b + ab^2 + a^2b - ab^2 + b^3$

$= a^3 + b^3$ (since $-a^2b + a^2b = 0$ and $+ab^2 - ab^2 = 0$)

Result: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

For systematic learning of polynomial factorisation see Algebra of Limits Overview.

Stepwise Application: Worked Example

Given: Expand $(x + 2)^3$.

Substitution: Let $a = x$, $b = 2$.

By the cube expansion, $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$:

$= x^3 + 3x^2\times2 + 3x\times(2^2) + (2^3)$

Simplification:

$= x^3 + 6x^2 + 12x + 8$

Final result: $(x + 2)^3 = x^3 + 6x^2 + 12x + 8$

More formula applications can be found in Progressions in Algebra.