Class 9 › Mathematics › Chapter

Exploring Algebraic Identities

Identities for squares, cubes and products, their geometric meaning, using them to calculate and factorise, simplifying rational expressions and solving quadratic equations by factorisation.

1. Identity vs Equation
Identity

True for all values

An algebraic identity is an equation that is true for every value of the variables in it. Both sides are just two forms of the same expression.

Equation

True for some values

An ordinary equation holds only for certain values (its solutions) and is false for other values.

  • Checking vs proving: testing a few numbers (including negatives and rational numbers) only suggests an identity; it is proved by expanding with the distributive property (or shown geometrically).
  • Geometric / algebra-tile models help visualise identities, but they are drawn for positive lengths; the algebraic proof shows the identity holds for all numbers.
  • Uses: expanding products, factorising expressions, simplifying calculations (squares, products, cubes) and simplifying rational expressions.
  • Proving number patterns: write the numbers algebraically (for example $n-1$, $n$, $n+1$ for three consecutive numbers), apply identities and simplify; the result then holds for every such set of numbers.
2. Square of a Sum: $(a+b)^2$
Identity 1 $$(a+b)^2 = a^2 + 2ab + b^2$$
a² ab ab b² ab ab Whole square of side (a + b): area (a + b)²
A square of side $a + b$ splits into a square $a^2$, a square $b^2$ and two rectangles $ab$.
Algebraic proof
$(a+b)^2 = (a+b)(a+b) = a(a+b) + b(a+b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2$.
This holds for all numbers $a, b$: positive, negative, rational (not only lengths).
  • Common error: $(a+b)^2 \neq a^2 + b^2$. The difference is $(a+b)^2 - (a^2 + b^2) = 2ab$.
  • Hence: $(a+b)^2 \gt a^2 + b^2$ when $ab \gt 0$ (same sign), $(a+b)^2 \lt a^2 + b^2$ when $ab \lt 0$ (opposite signs), and they are equal exactly when $ab = 0$ (one of them is $0$).
  • Expanding: treat each bracket term as a whole: $a$ and $b$ may be any expressions, e.g. products, fractions, or terms with coefficients (square each coefficient and variable, and the middle term is twice the product).
  • Squaring numbers: split the number into a convenient sum such as (round number) + (small number), then expand.
  • Factorising: an expression $a^2 + 2ab + b^2$ equals $(a+b)^2$ (see Section 10).
3. Square of a Difference: $(a-b)^2$
Identity 2 $$(a-b)^2 = a^2 - 2ab + b^2$$
Two ways to get it
Replace $b$ by $-b$ in Identity 1: $(a + (-b))^2 = a^2 + 2a(-b) + (-b)^2 = a^2 - 2ab + b^2$.
Geometric: take a square of side $a$, split $a$ into $(a-b)$ and $b$. Then $(a-b)^2 = a^2 - ab - b(a-b) = a^2 - 2ab + b^2$ (subtract the two rectangles $ab$ and $b(a-b)$ from the big square).
  • Only the middle term changes sign; $a^2$ and $b^2$ stay positive.
  • Use for squaring numbers slightly below a round number, and for factorising expressions of the form $a^2 - 2ab + b^2$.
  • Note: $(a-b)^2 = (b-a)^2$.
4. Square of Three Terms: $(a+b+c)^2$
Identity 3 $$(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$$
Derivation (substitution)
1. Put $d = b + c$: $(a + d)^2 = a^2 + 2ad + d^2$.
2. Replace $d$ back: $a^2 + 2a(b+c) + (b+c)^2 = a^2 + 2ab + 2ac + b^2 + 2bc + c^2$.
3. Geometrically, a square of side $a + b + c$ splits into three squares ($a^2, b^2, c^2$) and six rectangles (two each of $ab, bc, ca$).
  • Memory rule: sum of the squares of the three terms + twice the sum of the products of terms taken two at a time.
  • Negative terms: write each term with its sign (for example, treat $a - b + c$ as $a + (-b) + c$); the cross term sign is the product of the two signs.
  • Squaring numbers: split into three parts (hundreds + tens + units).
  • Factorising six-term expressions: if three terms are perfect squares and the other three are twice the products of their roots (with matching signs), the expression is $(a+b+c)^2$ with suitable signs.
  • It also gives $xy + yz + zx = \dfrac{(x+y+z)^2 - (x^2 + y^2 + z^2)}{2}$.
5. Difference of Squares
Identity 4 $$a^2 - b^2 = (a+b)(a-b)$$ Rearranged (Śhrīdharāchārya's method for squares) $$a^2 = (a+b)(a-b) + b^2$$
Proof
$(a+b)(a-b) = a^2 - ab + ba - b^2 = a^2 - b^2$ (the middle terms cancel).
Geometric: from a square of side $a$ remove a $b \times b$ corner; the remaining L-shape can be rearranged into a rectangle of sides $(a+b)$ and $(a-b)$.
  • Quick squares: choose $b$ so that $a + b$ or $a - b$ is a round number, then use $a^2 = (a+b)(a-b) + b^2$.
  • Quick products: two numbers equally spaced from a middle number $m$ (the numbers are $m - d$ and $m + d$) have product $m^2 - d^2$.
  • Squares ending in 5: $(10n + 5)^2 = 100\,n(n+1) + 25$, so the square is $n(n+1)$ followed by $25$.
  • Factorising: any difference of two perfect squares factorises as above (take out a common factor first if needed). It is the only way a binomial of the form $p^2 - q^2$ factorises; a sum of two squares $a^2 + b^2$ does not factorise over real numbers.
6. Product of Linear Factors and Factorising Quadratics
Identity 5 $$(x+a)(x+b) = x^2 + (a+b)x + ab$$ Identity 6 (general) $$(px + a)(qx + b) = pq\,x^2 + (pb + qa)\,x + ab$$
x² ax bx ab xa xb x² tilex tilesunit tiles
Algebra tiles: a rectangle with sides $x + a$ and $x + b$ is made of one $x^2$ tile, $a + b$ strips of $x$, and $ab$ unit tiles.
  • Algebra tiles: an $x^2$-tile (square of side $x$), $x$-tiles (rectangles $x \times 1$) and unit tiles ($1 \times 1$). Arranging all the tiles of an expression into a rectangle shows its factors (the rectangle's sides). Both multiplying and factorising are seen in the same picture.
  • Splitting the middle term: the $x$ term is split into two parts so that tiles form a rectangle; the split must be chosen correctly.
Factorising $x^2 + Sx + P$ (without tiles)
1. Compare with $x^2 + (a+b)x + ab$: find two numbers $a, b$ with $a + b = S$ and $ab = P$.
2. Then $x^2 + Sx + P = (x+a)(x+b)$.
3. Look at factor pairs of $P$ and test which pair adds to $S$.
Signs of $S$ and $P$Signs of $a$ and $b$
$P \gt 0$, $S \gt 0$Both positive
$P \gt 0$, $S \lt 0$Both negative
$P \lt 0$Opposite signs; the number with the larger absolute value has the sign of $S$
Factorising $px^2 + qx + r$ (split the middle term)
1. Find two numbers $m, n$ with $m + n = q$ and $mn = p \times r$ (product of first and last coefficients).
2. Rewrite $qx = mx + nx$.
3. Group the four terms in pairs, take the common factor from each pair, then take the common bracket out.
Why it works: in $(ax + b)(cx + d)$ the middle coefficient is $ad + bc$ and $(ad)(bc) = (ac)(bd)$.
  • Always check: expand your factors back to confirm.
  • If every term has a common factor, take it out first.
7. Cube of a Sum and of a Difference
Identity 7 and 8 $$(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$$ $$(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$$
Derivation
1. $(a+b)^3 = (a+b)(a^2 + 2ab + b^2) = a^3 + 3a^2b + 3ab^2 + b^3$ (distributive property).
2. Geometric: a cube of edge $a + b$ splits into two cubes (volumes $a^3$ and $b^3$) and six cuboids: three of size $a \times a \times b$ (volume $a^2b$ each) and three of size $a \times b \times b$ (volume $ab^2$ each).
3. Replace $b$ by $-b$ to get $(a-b)^3$. The four terms alternate in sign: $+, -, +, -$.
Useful equivalent forms $$(a+b)^3 = a^3 + b^3 + 3ab(a+b), \qquad (a-b)^3 = a^3 - b^3 - 3ab(a-b)$$
  • Recognising a cube: for an expression with four terms, check that the first and last terms are perfect cubes ($a^3$, $b^3$) and the middle terms are $3a^2b$ and $3ab^2$ (with the correct signs). Then it is $(a \pm b)^3$, which also gives the edge of a cube whose volume is that expression.
  • Cubing numbers: split the number into (round number) $\pm$ (small number).
  • A negative cube is negative: $(-t)^3 = -t^3$.
8. Sum and Difference of Cubes
Identity 9 and 10 $$x^3 - y^3 = (x-y)(x^2 + xy + y^2)$$ $$x^3 + y^3 = (x+y)(x^2 - xy + y^2)$$
Proof
$(x-y)(x^2 + xy + y^2) = x^3 + x^2y + xy^2 - x^2y - xy^2 - y^3 = x^3 - y^3$.
Replacing $y$ by $-y$ gives the sum-of-cubes identity.
  • Sign pattern (memory): the linear factor has the same sign as the original expression; in the quadratic factor the middle term has the opposite sign and the last term is always $+y^2$.
  • The quadratic factor $x^2 \pm xy + y^2$ does not factorise further over real numbers.
  • Pattern in $x^n - y^n$: $x - y$ is a factor of $x^2 - y^2$, $x^3 - y^3$, $x^4 - y^4$ (since $x^4 - y^4 = (x^2 - y^2)(x^2 + y^2)$), and in general of $x^n - y^n$ for every positive integer $n$: $$x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + \cdots + y^{n-1}).$$
  • Use it to factorise $p^3 \pm q^3$ where terms are perfect cubes (take out any common factor first), and to simplify rational expressions involving cubes.
9. The Identity $x^3 + y^3 + z^3 - 3xyz$
Identity 11 $$x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx)$$
Proof by expansion
1. Multiply each term of $(x+y+z)$ by each term of $(x^2 + y^2 + z^2 - xy - yz - zx)$.
2. All the mixed terms ($x^2y$, $xy^2$, $x^2z$, $xz^2$, $y^2z$, $yz^2$) cancel in pairs, and the three $xyz$ terms give $-3xyz$.
3. What remains is $x^3 + y^3 + z^3 - 3xyz$.
  • Special case: if $x + y + z = 0$, then $x^3 + y^3 + z^3 = 3xyz$.
  • Finding a sum of cubes from given data: if $x + y + z$, $xyz$ and $x^2 + y^2 + z^2$ are known, use Identity 3 to get $xy + yz + zx$ and then substitute in Identity 11.
  • Factorising an expression with three cubic terms and a "$-3\times$ product" term (with the right coefficients): treat the cube roots as $x, y, z$ and apply the identity.
  • Combining identities gives new identities (for example, $(a-b)^2(a+b) = (a^2 - 2ab + b^2)(a+b) = (a-b)(a^2 - b^2)$: both ways of grouping are correct).
10. Factorisation Strategy
Order of steps
1. Take out the common factor of all terms (including numerical factors).
2. Count the terms and match the pattern in the table.
3. Check that each factor cannot be factorised further ("factor completely"), then verify by expanding.
Number of termsPattern to look forFactorised form
2$a^2 - b^2$$(a+b)(a-b)$
2$a^3 - b^3$ / $a^3 + b^3$$(a \mp b)(a^2 \pm ab + b^2)$
3$a^2 \pm 2ab + b^2$ (perfect square)$(a \pm b)^2$
3$x^2 + Sx + P$$(x+a)(x+b)$, with $a + b = S$, $ab = P$
3$px^2 + qx + r$Split the middle term
4$a^3 \pm 3a^2b + 3ab^2 \pm b^3$$(a \pm b)^3$
4Three cubes and $-3xyz$ type term$(x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx)$
6Three squares and three cross terms$(a \pm b \pm c)^2$
  • Perfect square check for a trinomial: first and last terms are perfect squares and the middle term equals twice the product of their roots. If a fractional or irrational root appears, take a common factor out first.
  • Area and volume: if an area is given as an expression, its factors give possible length and breadth (answers are not unique); a volume expression factorised into three linear factors gives possible length, breadth and height.
  • Divisibility results: factorisation proves facts such as $n^3 - n = (n-1)\,n\,(n+1)$, a product of three consecutive integers, which is divisible by $6$.
11. Simplifying Rational Expressions
Rational expression: a fraction $\dfrac{P(x)}{Q(x)}$ of two polynomials, defined only when $Q(x) \neq 0$.
Steps
1. Factorise the numerator completely.
2. Factorise the denominator completely.
3. Cancel the common factors (a common factor of numerator and denominator).
4. State the assumption that the denominator (and so every cancelled factor) is not zero.
  • Cancel only factors, never separate terms joined by $+$ or $-$.
  • A factor can be cancelled because it is not zero, which follows from the denominator not being zero.
  • Use identities and middle-term splitting in the factorisation step (the same strategy as Section 10).
  • Powers such as $(a-b)^2$ in numerator and $(a-b)$ in denominator cancel one factor of $(a-b)$.
12. Solving Quadratic Equations by Factorisation
Zero-product property: if $p \times q = 0$, then $p = 0$ or $q = 0$ (or both).
Method
1. Form the equation from the problem (for example, area = length $\times$ breadth) and bring all terms to one side so the other side is $0$: $ax^2 + bx + c = 0$.
2. Factorise the left side (split the middle term if needed).
3. Put each factor equal to $0$ and solve.
4. Reject solutions that do not make sense in the context (for example, a negative length) and state the final answer with units.
  • Modelling areas: a rectangle made from an $x^2$ square, strips and unit squares has area $x^2 + (\text{number of strips})x + (\text{number of units})$; factorise it to get the dimensions.
  • A product equal to a non-zero number does not let you equate each factor to that number; first make one side zero.
13. All Identities and Quick Revision
No.Identity
1$(x+y)^2 = x^2 + 2xy + y^2$
2$(x-y)^2 = x^2 - 2xy + y^2$
3$(x+y+z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$
4$(x+y)(x-y) = x^2 - y^2$
5$(x+a)(x+b) = x^2 + (a+b)x + ab$
6$(ax+b)(cx+d) = acx^2 + (ad+bc)x + bd$
7$x^3 - y^3 = (x-y)(x^2 + xy + y^2)$
8$x^3 + y^3 = (x+y)(x^2 - xy + y^2)$
9$(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$
10$(x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3$
11$x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - xz - yz)$
Calculations

Which identity to use

Number near a round number: $(a \pm b)^2$ or $(a \pm b)^3$. Numbers equally spaced about a middle value: $(a+b)(a-b)$. Number with three parts: $(a+b+c)^2$.

Check

Verify by expanding

Multiply out the factors you found; you must get the original expression.

Common mistakes:
• Writing $(a+b)^2 = a^2 + b^2$ or $(a+b)^3 = a^3 + b^3$.
• Forgetting the middle term $2ab$, or giving it the wrong sign in $(a-b)^2$.
• Not alternating signs in $(a-b)^3$.
• Mixing up the signs in $x^3 \pm y^3$ factors.
• Cancelling terms instead of factors in a rational expression.
• Equating factors to a non-zero number while solving a quadratic.
• Forgetting the common factor before applying an identity.