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CDS Mathematics · Formula sheet

Algebraic Identities and Simplification formulas

21 formulas and 21 common traps for CDS Mathematics Algebraic Identities and Simplification, grouped by subtopic.

Full notes with worked examples

Squares and Cubes of a Binomial

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Everything from a + b and ab

Cube of a sum

a3+b3=(a+b)3−3ab(a+b)a^3 + b^3 = (a + b)^3 - 3ab(a + b)

Spotting an expanded cube

Cube of a difference

(a−b)3=a3−3a2b+3ab2−b3(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3

Difference of squares, sum and difference of cubes

Sum and difference of cubes

a3±b3=(a±b)(a2∓ab+b2)a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)

Common traps

The correction term is 3ab(a + b)

(a+b)3(a + b)^3 is a3+b3+3ab(a+b)a^3 + b^3 + 3ab(a + b), not a3+b3+3aba^3 + b^3 + 3ab. Forgetting the factor (a+b)(a + b) gives a wrong answer that is often printed as an option.

Check the middle coefficients

27x3+54x2y+36xy2+8y327x^3 + 54x^2y + 36xy^2 + 8y^3 is (3x+2y)3(3x + 2y)^3 because 3(3x)2(2y)=54x2y3(3x)^2(2y) = 54x^2y. If a middle term does not match, the four terms are not a cube and the shortcut fails.

The signs pair up opposite

a3+b3a^3 + b^3 goes with a2−ab+b2a^2 - ab + b^2, and a3−b3a^3 - b^3 with a2+ab+b2a^2 + ab + b^2. Read the sign of the middle term in the denominator first; it tells you which cube to look for on top.

Reciprocal Sums x ± 1/x

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Squaring up and down the ladder

Squaring a reciprocal sum

(x±1x)2=x2+1x2±2\left(x \pm \tfrac1x\right)^2 = x^2 + \tfrac{1}{x^2} \pm 2

Cubes and higher powers

Cubes of a reciprocal sum

x3+1x3=u3−3u,x3−1x3=v3+3vx^3 + \tfrac{1}{x^3} = u^3 - 3u, \qquad x^3 - \tfrac{1}{x^3} = v^3 + 3v

Common traps

x − 1/x has two signs

Coming down from x2+1x2x^2 + \dfrac{1}{x^2}, x−1xx - \dfrac1x is ±\pm a square root, and x>0x > 0 does not decide the sign (x=2x = 2 and x=12x = \dfrac12 both have x>0x > 0). If the options list only one sign, that is the intended value; if both appear, look for a condition like x>1x > 1.

Plus 3v for the difference, minus 3u for the sum

(x−1x)3=x3−1x3−3(x−1x)\left(x - \dfrac1x\right)^3 = x^3 - \dfrac{1}{x^3} - 3\left(x - \dfrac1x\right), so the difference of cubes ADDS 3v3v. Using −3v-3v by analogy with the sum gives a value that is usually one of the options.

Sums and Products of Three Variables

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The square of a + b + c

Square of a sum of three

(a+b+c)2=a2+b2+c2+2(ab+bc+ca)(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)

Expanding a product of brackets

Cubic from its roots

(x−a)(x−b)(x−c)=x3−∑a x2+∑ab x−abc(x - a)(x - b)(x - c) = x^3 - \textstyle\sum a\, x^2 + \sum ab\, x - abc

When 2s = a + b + c

Semi-perimeter pieces

(s−a)+(s−b)+(s−c)=s(s - a) + (s - b) + (s - c) = s

Common traps

A square root gives two signs

From (a+b+c)2=64(a + b + c)^2 = 64, a+b+c=±8a + b + c = \pm 8. Keep both unless the question says the numbers are positive.

The signs alternate

In (x−a)(x−b)(x−c)(x - a)(x - b)(x - c) the coefficients go +,−,+,−+, -, +, -: the xx term is +∑ab+\sum ab and the constant is −abc-abc. With (x+a)(x+b)(x+c)(x + a)(x + b)(x + c) every sign is ++.

s is half the perimeter

2s=a+b+c2s = a + b + c, so s−a=b+c−a2s - a = \dfrac{b + c - a}{2}, not b+c−ab + c - a. Substituting s=a+b+cs = a + b + c doubles every term.

The Cube Identity and a + b + c = 0

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When three numbers add to zero

Zero sum

a+b+c=0  ⇒  a3+b3+c3=3abca + b + c = 0 \;\Rightarrow\; a^3 + b^3 + c^3 = 3abc

The full factorisation

Cube identity

a3+b3+c3−3abc=(a+b+c)[(a+b+c)2−3(ab+bc+ca)]a^3 + b^3 + c^3 - 3abc = (a + b + c)\left[(a + b + c)^2 - 3(ab + bc + ca)\right]

Common traps

Check that the three really add to zero

(x−y)(x - y), (y−z)(y - z), (z−x)(z - x) add to 00; (x−y)(x - y), (y−z)(y - z), (x−z)(x - z) do not. Read the third bracket's order before using the shortcut.

Zero does not mean a + b + c = 0

a3+b3+c3−3abc=0a^3 + b^3 + c^3 - 3abc = 0 also holds when a=b=ca = b = c. A statement claiming it forces the numbers to be equal is not enough; (1,−1,0)(1, -1, 0) satisfies it with unequal numbers.

Sums of Squares and Least Values

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Sums of squares that vanish

Sum of squared differences

a2+b2+c2−ab−bc−ca=12[(a−b)2+(b−c)2+(c−a)2]a^2 + b^2 + c^2 - ab - bc - ca = \tfrac12\left[(a - b)^2 + (b - c)^2 + (c - a)^2\right]

Least values: t + 1/t is at least 2

AM–GM

pt+qt≥2pq(p,q,t>0)pt + \dfrac{q}{t} \ge 2\sqrt{pq} \quad (p, q, t > 0)

Which is larger?

Mean of squares against square of mean

a2+b22−(a+b2)2=(a−b)24≥0\dfrac{a^2 + b^2}{2} - \left(\dfrac{a + b}{2}\right)^2 = \dfrac{(a - b)^2}{4} \ge 0

Common traps

Read what the stem already gives

If the stem says aa, bb, cc are distinct, then a2+b2+c2−ab−bc−caa^2 + b^2 + c^2 - ab - bc - ca is already known to be positive, and no statement is needed. Data-sufficiency items set this on purpose.

Only for positive values

For negative tt, t+1t≤−2t + \dfrac1t \le -2, so the expression has no least value on all reals. The condition x>0x > 0 in the stem is what makes the answer exist.

Greater or greater-or-equal?

A difference like x2y2(x2−y2)2x^2y^2(x^2 - y^2)^2 is never negative but is zero when x=−yx = -y. If the question asks 'always greater', that single case answers no.

Conditional Identities

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Rewrite each piece with the condition

A typical rewrite

ab+bc+ca=0  ⇒  a2−bc=a(a+b+c)ab + bc + ca = 0 \;\Rightarrow\; a^2 - bc = a(a + b + c)

Equal ratios

Adding equal ratios

ab=cd=ef=k  ⇒  k=a+c+eb+d+f\dfrac ab = \dfrac cd = \dfrac ef = k \;\Rightarrow\; k = \dfrac{a + c + e}{b + d + f}

Factor first, then substitute

Divide out the known factor

x2+xy−2y2=(x+2y)(x−y)x^2 + xy - 2y^2 = (x + 2y)(x - y)

Common traps

Test with numbers that fit

When ab+bc+ca=0ab + bc + ca = 0, a=b=1a = b = 1 does not fit unless c=−12c = -\dfrac12. Build the test triple from the condition, then evaluate; a single wrong-looking value rules out options quickly.

Do not drop the zero-sum case

Adding numerators and denominators divides by their sum. When that sum can be zero, the ratio takes a second value, and the options usually include both the full answer and the half answer.

Dividing by something that could be zero

From x2(m−1)=mab(m−1)x^2(m - 1) = mab(m - 1) you may cancel m−1m - 1 only if m≠1m \ne 1. From (x−y)k=(y−z)k(x - y)k = (y - z)k you may not conclude x−y=y−zx - y = y - z unless k≠0k \ne 0 — and sometimes k=0k = 0 is the answer the question wants.

Rational Algebraic Expressions

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Factorise and cancel

Cancel a common factor

(x−1)(x2+x+1)x2+x+1=x−1\dfrac{(x - 1)(x^2 + x + 1)}{x^2 + x + 1} = x - 1

Combining fractions

Two fractions

1x−1−1x+1=2x2−1\dfrac{1}{x - 1} - \dfrac{1}{x + 1} = \dfrac{2}{x^2 - 1}

Equations with fractions

Excluded values

P(x)Q(x)=0  ⇒  P(x)=0, Q(x)≠0\dfrac{P(x)}{Q(x)} = 0 \;\Rightarrow\; P(x) = 0,\ Q(x) \ne 0

Common traps

Cancel factors, not terms

In x2+3xx+3\dfrac{x^2 + 3x}{x + 3} you may cancel (x+3)(x + 3) after writing the top as x(x+3)x(x + 3). You may not cross out the 3x3x against the 33.

Order of subtraction

'What should be added to PP to get QQ' is Q−PQ - P, not P−QP - Q. The reversed answer differs only in sign and is usually printed.

Check the denominators

A root that makes an original denominator zero is not a solution, even though it solves the cleared polynomial. And x=0x = 0 often satisfies a symmetric equation trivially; the question usually wants the other roots.

Cyclic Sums and Factors

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Three cyclic fraction sums

The key cyclic sum

a2(a−b)(a−c)+b2(b−c)(b−a)+c2(c−a)(c−b)=1\dfrac{a^2}{(a - b)(a - c)} + \dfrac{b^2}{(b - c)(b - a)} + \dfrac{c^2}{(c - a)(c - b)} = 1

Factors of cyclic expressions

The basic cyclic factorisation

a(b2−c2)+b(c2−a2)+c(a2−b2)=(a−b)(b−c)(c−a)a(b^2 - c^2) + b(c^2 - a^2) + c(a^2 - b^2) = (a - b)(b - c)(c - a)

Common traps

Watch the order of the factors

Swapping one factor, (a−b)(a - b) for (b−a)(b - a), flips the sign of that term. A sum that 'should' be 11 can be −1-1 if the question writes the denominators differently — check with numbers.

One test disproves; it does not prove

A single triple giving a non-zero value shows a factor is absent. A triple giving zero shows nothing by itself — use the vanishing-when-equal argument to prove a factor is present.

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