Why is "Treating $x^{2/3}$ as $\frac{2}{3} x$" a common mistake in Indices and Surds?

Why it happens: Confusion of coefficient with power. How to avoid it: x^2/3 means take the cube root then square (or square then cube root). NOT a coefficient. Use the formula x^a/b = √[b]x^a. Source: 0606 Examiner Reports 2022-2024

Why is "Adding bases when adding powers" a common mistake in Indices and Surds?

Why it happens: Confusion: x^a · x^b = x^a+b adds powers, but you might think bases should be added. How to avoid it: Bases STAY THE SAME. Only powers add (when multiplying). x^2 + x^3 x^5. Don't combine when adding.

Why is "Thinking $x^{-1} = -x$" a common mistake in Indices and Surds?

Why it happens: Mixing negative powers with negation. How to avoid it: x^-1 = 1/x, NOT -x. Negative power means RECIPROCAL.

Why is "Using the wrong sign in the conjugate" a common mistake in Indices and Surds?

Why it happens: Sign confusion. How to avoid it: If denominator is a + b√(c), conjugate is a - b√(c) (FLIP the sign). Multiplying gives a^2 - b^2 c — no surds.

Indices and SurdsCambridge IGCSE Additional Mathematics 0606

Indices and Surds

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Short Study Notes — Indices and Surds

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Indices and Surds Study Notes — Cambridge IGCSE Additional Mathematics 0606 (2025-2027 syllabus)

Laws of indices: products, quotients, powers, fractional and negative powers. Manipulating surds: simplifying, rationalising the denominator, conjugates. Solving simple exponential equations.

What you’ll learn

Mapped to the Cambridge IGCSE 0606 syllabus (2025-2027).

1.4.1 — Apply the laws of indices.
1.4.2 — Simplify expressions with fractional and negative powers.
1.4.3 — Simplify and combine surds.
1.4.4 — Rationalise denominators using conjugates.

Laws of indices

Six rules; learn them as a set.

Standard laws (assume same base $x$ , real powers):

Law	Form
Product	$x^a \cdot x^b = x^{a+b}$
Quotient	$\dfrac{x^a}{x^b} = x^{a-b}$
Power of a power	$(x^a)^b = x^{ab}$
Zero power	$x^0 = 1$ (for $x \ne 0$ )
Negative power	$x^{-a} = \dfrac{1}{x^a}$
Fractional power	$x^{a/b} = \sqrt[b]{x^a} = (\sqrt[b]{x})^a$

Two-base rule. $(xy)^a = x^a y^a$ — distributes.

Worked walkthrough. Simplify $\dfrac{8x^4 \cdot 2x^{-3}}{4x^{-2}}$ .

Numerator: $8 \cdot 2 = 16$ ; $x^{4 + (-3)} = x^1$ . So $16x$ .
Divide: $\dfrac{16x}{4x^{-2}} = 4 \cdot x^{1 - (-2)} = 4x^3$ .

The six index laws underpin every algebraic manipulation in Additional Mathematics — same base, multiply means add powers; divide means subtract; fractional powers become roots.

Cambridge tip. Match the index law to the operation: same base, multiplying → ADD powers; dividing → SUBTRACT.

Six index laws — memorise.
Same base, multiply → add powers.
Negative powers = reciprocals.
Fractional powers = roots.

See the full worked example for indices and surds →

Fractional and negative powers

$x^{a/b} = \sqrt[b]{x^a}$ and $x^{-a} = \frac{1}{x^a}$ .

Fractional powers. $x^{a/b}$ means take the $b$ th root of $x^a$ (or equivalently, the $b$ th root of $x$ raised to the power $a$ ).

Examples:

$x^{1/2} = \sqrt{x}$ .
$x^{1/3} = \sqrt[3]{x}$ .
$x^{2/3} = \sqrt[3]{x^2} = (\sqrt[3]{x})^2$ .

Negative powers. $x^{-a} = \dfrac{1}{x^a}$ . Think 'reciprocal'.

Examples:

$x^{-1} = \dfrac{1}{x}$ .
$x^{-2} = \dfrac{1}{x^2}$ .
$x^{-1/2} = \dfrac{1}{\sqrt{x}}$ .

Solving simple equations. $x^{2/3} = 9$ .

Raise both sides to the reciprocal power: $\left(x^{2/3}\right)^{3/2} = 9^{3/2}$ .
Left side becomes $x^1 = x$ .
Right side: $9^{3/2} = (\sqrt{9})^3 = 27$ .
So $x = 27$ .

Cambridge tip. Always state both the simplified form AND verify by substitution where possible.

$x^{a/b} = \sqrt[b]{x^a}$ .
$x^{-a} = \frac{1}{x^a}$ .
Solve $x^{a/b} = c$ by raising to the reciprocal power.

See the full worked example for indices and surds →

Surds

Irrational roots. Simplify by pulling out square factors.

Surd = an irrational root that cannot be reduced to a rational number — e.g. $\sqrt{2}$ , $\sqrt{18}$ , $\sqrt[3]{5}$ .

Simplifying. Pull out the largest perfect square factor under the root.

$\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}$ .
$\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}$ .

Combining.

$\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$ (when $a, b \ge 0$ ).
$\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}$ .
Adding / subtracting: ONLY combine LIKE surds. $2\sqrt{3} + 5\sqrt{3} = 7\sqrt{3}$ , but $2\sqrt{3} + 5\sqrt{2}$ does NOT simplify.

Caution. $\sqrt{a + b} \ne \sqrt{a} + \sqrt{b}$ . This is a common error.

Cambridge tip. Look for perfect-square factors (4, 9, 16, 25, ...) under any surd. If pulled out, the surd becomes simpler.

Pull out perfect-square factors.
Like surds add; unlike don't.
$\sqrt{a+b} \ne \sqrt{a} + \sqrt{b}$ .

Rationalising the denominator

Multiply by the conjugate to remove surds from the denominator.

Why rationalise. Mathematical convention prefers no surds in the denominator. Easier to compare and add fractions.

Simple case: $\dfrac{a}{\sqrt{b}}$ . Multiply top and bottom by $\sqrt{b}$ : $\dfrac{a}{\sqrt{b}} \cdot \dfrac{\sqrt{b}}{\sqrt{b}} = \dfrac{a\sqrt{b}}{b}.$

Compound case: $\dfrac{a}{p + q\sqrt{r}}$ . Multiply top and bottom by the CONJUGATE $(p - q\sqrt{r})$ : $\dfrac{a}{p + q\sqrt{r}} \cdot \dfrac{p - q\sqrt{r}}{p - q\sqrt{r}} = \dfrac{a(p - q\sqrt{r})}{p^2 - q^2 r}.$

The denominator becomes RATIONAL because of the difference-of-squares pattern.

Worked walkthrough. Simplify $\dfrac{6}{2 + \sqrt{3}}$ .

Multiply by conjugate: $\dfrac{6}{2 + \sqrt{3}} \cdot \dfrac{2 - \sqrt{3}}{2 - \sqrt{3}} = \dfrac{6(2 - \sqrt{3})}{4 - 3} = 6(2 - \sqrt{3}) = 12 - 6\sqrt{3}$ .

Cambridge tip. When the denominator is $p \pm q\sqrt{r}$ , the conjugate is $p \mp q\sqrt{r}$ . Multiplying gives $p^2 - q^2 r$ — no surds.

Conjugate flips the sign of the surd term.
$(a + b\sqrt{c})(a - b\sqrt{c}) = a^2 - b^2 c$ .
Always rationalise the denominator in final answers.

See the full worked example for indices and surds →

How it’s examined

Indices and surds appear on most Paper 1s. Most-tested: simplify with index laws (4-5 marks), rationalise denominator (4-5 marks), solve fractional-power equations (3-5 marks).

Sources: Cambridge IGCSE Additional Mathematics 0606 syllabus (2025-2027); 0606 Examiner Reports 2022-2024; 0606/12 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-09.

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Step-by-step worked examples — Indices and Surds

Step-by-step solutions to past-paper-style questions on indices and surds, written exactly the way a tutor would explain them at the board.

1Apply the laws of indices (5 marks)
Extended• Adapted from 0606/12 May/Jun 2024 Q2• indices
Question
Simplify $\dfrac{8x^4 \cdot 2x^{-3}}{4x^{-2}}$ . (5 marks)
Step-by-step solution
1. Step 1
  Multiply numerator (2 marks). $8x^4 \cdot 2x^{-3} = 16x^{4 + (-3)} = 16x$ .
2. Step 2
  Divide (2 marks). $\dfrac{16x}{4x^{-2}} = 4 \cdot x^{1 - (-2)} = 4x^3$ .
3. Step 3
  Final (1 mark). Answer: $4x^3$ .
Answer
$4x^3$ .
2Rationalise the denominator (4 marks)
Extended• surds, rationalise
Question
Simplify $\dfrac{6}{2 + \sqrt{3}}$ in the form $a + b\sqrt{3}$ . (4 marks)
Step-by-step solution
1. Step 1
  Multiply by the conjugate (2 marks). Multiply numerator and denominator by $(2 - \sqrt{3})$ . $\dfrac{6}{2 + \sqrt{3}} \cdot \dfrac{2 - \sqrt{3}}{2 - \sqrt{3}} = \dfrac{6(2 - \sqrt{3})}{(2)^2 - (\sqrt{3})^2}$ .
2. Step 2
  Simplify denominator (1 mark). $(2)^2 - (\sqrt{3})^2 = 4 - 3 = 1$ .
3. Step 3
  Result (1 mark). $\dfrac{6(2 - \sqrt{3})}{1} = 12 - 6\sqrt{3}$ .
Answer
$12 - 6\sqrt{3}$ , so $a = 12$ and $b = -6$ .
3Solve a fractional-power equation (4 marks)
Extended• indices, fractional-power
Question
Solve $x^{2/3} = 9$ . (4 marks)
Step-by-step solution
1. Step 1
  Raise both sides to the reciprocal power (2 marks). $\left(x^{2/3}\right)^{3/2} = 9^{3/2}$ .
2. Step 2
  Simplify left side (1 mark). $x^{(2/3)(3/2)} = x^1 = x$ .
3. Step 3
  Compute right side (1 mark). $9^{3/2} = (\sqrt{9})^3 = 3^3 = 27$ . So $x = 27$ .
Answer
$x = 27$ .
Examiner tip
Check by substituting back: $27^{2/3} = (27^{1/3})^2 = 3^2 = 9$ . ✓

Key Formulae — Indices and Surds

The formulae you need to memorise for indices and surds on the Cambridge IGCSE 0606 paper, with every variable defined in plain English and a note on when to use it.

Product law
$x^a \cdot x^b = x^{a+b}$
$x$
Any base (real number, positive when fractional powers used)
$a, b$
Powers (real numbers)
When to use
Multiplying powers of the same base.
Quotient law
$\frac{x^a}{x^b} = x^{a-b}$
$x$
Any base
$a, b$
Powers
When to use
Dividing powers of the same base.
Power of a power
$(x^a)^b = x^{ab}$
$x, a, b$
Base and powers
When to use
Raising a power to another power.
Zero power
$x^0 = 1$
$x$
Any non-zero base
When to use
Anything (non-zero) to the power 0 equals 1.
Negative power
$x^{-a} = \frac{1}{x^a}$
$x, a$
Base and power
When to use
Convert negative powers to reciprocals.
Fractional power
$x^{a/b} = \sqrt[b]{x^a} = \left(\sqrt[b]{x}\right)^a$
$x$
Base (positive when $b$ is even)
$a, b$
Integers, $b > 0$
When to use
Convert between fractional powers and roots.
Surd product
$\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$
$a, b$
Non-negative reals
When to use
Combining surd products.
Conjugate (rationalising)
$(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$
$a, b$
Non-negative reals
When to use
Rationalising a denominator of the form $\sqrt{a} + \sqrt{b}$ — multiply numerator and denominator by the conjugate.

Key Definitions and Keywords — Indices and Surds

Definitions to memorise and the exact keywords mark schemes credit for indices and surds answers — sharpened from recent examiner reports for the 2026 0606 sitting.

Index (exponent)
Examiner keyword
The power to which a base is raised. In $x^a$ , $a$ is the index.
Surd
Examiner keyword
An irrational root that cannot be simplified to a rational number — e.g. $\sqrt{2}$ , $\sqrt[3]{5}$ .
Rationalise the denominator
Examiner keyword
Eliminate surds from a denominator by multiplying numerator and denominator by an appropriate factor (often the conjugate).
Conjugate
Examiner keyword
The conjugate of $a + b\sqrt{c}$ is $a - b\sqrt{c}$ . Multiplying gives $a^2 - b^2 c$ — rational.

Common Mistakes and Misconceptions — Indices and Surds

The traps other students keep falling into on indices and surds questions — taken from recent Cambridge IGCSE 0606 examiner reports and mark schemes — and how to avoid them.

✕Treating $x^{2/3}$ as $\frac{2}{3} x$
0606 Examiner Reports 2022-2024
Why it happens
Confusion of coefficient with power.
How to avoid it
$x^{2/3}$ means take the cube root then square (or square then cube root). NOT a coefficient. Use the formula $x^{a/b} = \sqrt[b]{x^a}$ .
✕Adding bases when adding powers
Why it happens
Confusion: $x^a \cdot x^b = x^{a+b}$ adds powers, but you might think bases should be added.
How to avoid it
Bases STAY THE SAME. Only powers add (when multiplying). $x^2 + x^3 \ne x^5$ . Don't combine when adding.
✕Thinking $x^{-1} = -x$
Why it happens
Mixing negative powers with negation.
How to avoid it
$x^{-1} = \frac{1}{x}$ , NOT $-x$ . Negative power means RECIPROCAL.
✕Using the wrong sign in the conjugate
Why it happens
Sign confusion.
How to avoid it
If denominator is $a + b\sqrt{c}$ , conjugate is $a - b\sqrt{c}$ (FLIP the sign). Multiplying gives $a^2 - b^2 c$ — no surds.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

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Indices and Surds

Short Study Notes in the form of Slides

Short Study Notes — Indices and Surds

Detailed Notes

What you’ll learn

How it’s examined

1Apply the laws of indices (5 marks)

2Rationalise the denominator (4 marks)

3Solve a fractional-power equation (4 marks)

Product law

Quotient law

Power of a power

Zero power

Negative power

Fractional power

Surd product

Conjugate (rationalising)

Index (exponent)

Surd

Rationalise the denominator

Conjugate

✕Treating x2/3x^{2/3}x2/3 as 23x\frac{2}{3} x32​x

✕Adding bases when adding powers

✕Thinking x−1=−xx^{-1} = -xx−1=−x

✕Using the wrong sign in the conjugate

Practice questions

Past paper style quiz

✕Treating $x^{2/3}$ as $\frac{2}{3} x$

✕Thinking $x^{-1} = -x$