What are powers and how are they used in mathematics?

In mathematics, powers refer to the operation of raising a number to an exponent, which indicates how many times the number is multiplied by itself. For example, 3^4 means 3 multiplied by itself 4 times, resulting in 81. Powers are used to simplify expressions and solve equations in the Edexcel IGCSE Mathematics curriculum.

How do you calculate the square root of a number?

The square root of a number is a value that, when multiplied by itself, gives the original number. For example, the square root of 16 is 4, since 4 x 4 = 16. To calculate square roots, you can use a calculator or estimate by finding the closest perfect squares. Understanding square roots is essential in the Numbers and the Number System topic of Edexcel IGCSE Mathematics.

What is the difference between a square and a cube in terms of powers?

A square refers to raising a number to the power of 2, meaning the number is multiplied by itself once (e.g., 5^2 = 25). A cube refers to raising a number to the power of 3, meaning the number is multiplied by itself twice (e.g., 2^3 = 8). Both concepts are fundamental in the study of powers and roots in the Edexcel IGCSE Mathematics curriculum.

How do you simplify expressions with powers?

To simplify expressions with powers, you apply the laws of exponents, such as the product of powers rule (a^m * a^n = a^(m+n)), the power of a power rule ((a^m)^n = a^(m*n)), and the quotient of powers rule (a^m / a^n = a^(m-n)). These rules help in simplifying complex expressions and are a key part of the Edexcel IGCSE Mathematics syllabus.

Why are roots important in solving equations?

Roots are important in solving equations because they allow us to find values that satisfy the equation. For example, solving x^2 = 9 involves finding the square root of 9, which gives x = 3 or x = -3. Understanding how to work with roots is crucial for solving quadratic equations and other mathematical problems in the Edexcel IGCSE Mathematics curriculum.

Why is "Adding indices when multiplying different bases" a common mistake in Powers and Roots?

Why it happens: Students apply x^a × x^b = x^a+b to different bases too. How to avoid it: Index laws ONLY work for the SAME base. x^3 × y^4 doesn't simplify; x^3 × x^4 = x^7. Source: 4MA1/1H — examiner reports 2022-2024

Why is "Writing $x^{0} = 0$ or $x^{0} = x$" a common mistake in Powers and Roots?

Why it happens: Zero suggests 'nothing'; some students think 0 powers cancel. How to avoid it: Memorise: x^0 = 1 for any non-zero x. Reason: x^nx^n = x^n-n = x^0, and any number divided by itself is 1.

Why is "Writing $x^{-n} = -x^{n}$" a common mistake in Powers and Roots?

Why it happens: Students confuse 'negative index' with 'negative number'. How to avoid it: Negative index means RECIPROCAL: x^-n = 1/x^n. The base stays positive (if it was positive). Source: 4MA1/1H May/Jun 2023 — examiner report Q9

Why is "Computing $27^{2/3}$ as $(27^{2})^{1/3}$ on a non-calculator paper" a common mistake in Powers and Roots?

Why it happens: Both methods give the same answer, but 27^2 = 729 is harder to take cube root of mentally. How to avoid it: ALWAYS take the root FIRST: 27^1/3 = 3, then square to get 9. Smaller numbers, less error. Source: 4MA1/1H — examiner reports 2022-2024

Why is "Forgetting to apply the power to constants in brackets" a common mistake in Powers and Roots?

Why it happens: Students apply the power only to the variable: (2x)^3 = 2x^3 (wrong). How to avoid it: (ab)^n = a^n · b^n. So (2x)^3 = 2^3 · x^3 = 8x^3. The power applies to EVERYTHING in the bracket. Source: 4MA1/1H Jan 2024 — examiner report Q11

Why is "Writing $(x+y)^{2} = x^{2} + y^{2}$" a common mistake in Powers and Roots?

Why it happens: Confusing with the product rule. How to avoid it: Index laws DON'T distribute over addition. (x+y)^2 = x^2 + 2xy + y^2 (use FOIL).

Numbers and the Number SystemPearson Edexcel IGCSE Mathematics 4MA1 Higher

Powers and Roots

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Powers and Roots Study Notes — Edexcel IGCSE 4MA1 Higher Tier (2026 onwards)

Index laws power algebra, surds, standard form, and exponential growth. Master the rules and you unlock half the Number topic.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

1.6 — Use index laws (product, quotient, power, zero, negative, fractional).
1.6 — Evaluate fractional and negative indices.
1.6 — Use square roots, cube roots and other roots.
1.6 — Estimate roots of non-perfect powers.

The seven index laws

Memorise these seven. Every algebra and surd question on Edexcel uses at least one.

The index laws apply to ANY base $x$ (with $x \ne 0$ where division is involved):

Law	Rule
Product	$x^{a} \times x^{b} = x^{a+b}$
Quotient	$\dfrac{x^{a}}{x^{b}} = x^{a-b}$
Power of a power	$(x^{a})^{b} = x^{ab}$
Power of a product	$(xy)^{n} = x^{n} y^{n}$
Power of a quotient	$\left(\dfrac{x}{y}\right)^{n} = \dfrac{x^{n}}{y^{n}}$
Zero index	$x^{0} = 1$ ( $x \ne 0$ )
Negative index	$x^{-n} = \dfrac{1}{x^{n}}$

These rules ONLY apply to the SAME base. $x^{3} \times y^{4}$ does NOT simplify; $x^{3} \times x^{4} = x^{7}$ does.

Examples.

Expression	Simplified
$2^{5} \times 2^{3}$	$2^{5+3} = 2^{8} = 256$
$\dfrac{x^{7}}{x^{3}}$	$x^{7-3} = x^{4}$
$(x^{4})^{3}$	$x^{4 \times 3} = x^{12}$
$(2x)^{3}$	$2^{3} x^{3} = 8x^{3}$
$5^{0}$	$1$
$3^{-2}$	$\dfrac{1}{3^{2}} = \dfrac{1}{9}$

Worked qualitative. Why does $x^{a} \times x^{b} = x^{a+b}$ ?

$x^{a} = x \cdot x \cdot \ldots \cdot x$ ( $a$ times).
$x^{b} = x \cdot x \cdot \ldots \cdot x$ ( $b$ times).
Multiplying: $x \cdot x \cdot \ldots \cdot x$ ( $a + b$ times) $= x^{a+b}$ .
Counting matters more than memorising — if you know what $x^{n}$ means, the product rule is automatic.

Multiplying powers of the same base just stacks the factors — so the indices add.

Edexcel tip. Show every step in algebra simplification questions. Mark schemes have separate marks for each application of an index law.

Same base only.
Product → ADD indices.
Quotient → SUBTRACT.
Power of power → MULTIPLY.
Zero → 1.
Negative → reciprocal.

Fractional and negative indices

Fractional = root + power. Negative = reciprocal. Always tackle the trickier bit first.

Fractional indices combine roots with powers.

$x^{m/n} = \sqrt[n]{x^{m}} = (\sqrt[n]{x})^{m}$

For non-calculator papers, ALWAYS take the root FIRST (smaller numbers, less arithmetic error).

Denominator gives the root, numerator gives the power — take the root first.

Examples:

Expression	Step	Result
$8^{1/3}$	cube root	$2$
$16^{1/2}$	square root	$4$
$8^{2/3}$	cube root, then square	$(2)^{2} = 4$
$27^{2/3}$	cube root, then square	$(3)^{2} = 9$
$64^{5/6}$	sixth root, then fifth power	$(2)^{5} = 32$
$32^{4/5}$	fifth root, then fourth power	$(2)^{4} = 16$

Negative indices mean RECIPROCAL.

$x^{-n} = \dfrac{1}{x^{n}}$

For fractions: flip the fraction, then apply the positive power.

$\left(\dfrac{a}{b}\right)^{-n} = \left(\dfrac{b}{a}\right)^{n}$

Examples:

Expression	Step	Result
$2^{-3}$	reciprocal	$\dfrac{1}{2^{3}} = \dfrac{1}{8}$
$5^{-2}$	reciprocal	$\dfrac{1}{25}$
$\left(\dfrac{2}{3}\right)^{-2}$	flip, then square	$\left(\dfrac{3}{2}\right)^{2} = \dfrac{9}{4}$
$\left(\dfrac{1}{4}\right)^{-3}$	flip, then cube	$4^{3} = 64$

Combining negative AND fractional.

$8^{-2/3}$ :

Negative → reciprocal: $\dfrac{1}{8^{2/3}}$ .
Fractional → root first: $8^{1/3} = 2$ , then $2^{2} = 4$ .
Final: $\dfrac{1}{4}$ .

Worked qualitative. Why does $x^{1/2}$ mean $\sqrt{x}$ ?

We want $x^{1/2}$ to follow the index laws.
$x^{1/2} \times x^{1/2} = x^{1/2 + 1/2} = x^{1}$ .
So $x^{1/2}$ is something that, multiplied by itself, gives $x$ .
That's the definition of $\sqrt{x}$ .

Edexcel tip. Memorise common fractional powers. $8^{1/3} = 2$ , $27^{1/3} = 3$ , $64^{1/3} = 4$ , $125^{1/3} = 5$ . $16^{1/4} = 2$ , $81^{1/4} = 3$ . $32^{1/5} = 2$ , $243^{1/5} = 3$ .

Fractional: root first, then power.
$x^{1/n} = \sqrt[n]{x}$ .
Negative: reciprocal.
$x^{-n} = 1/x^{n}$ .
Combined: handle one at a time.

Square, cube and other roots

Inverse of powering. Used for surds, Pythagoras, areas, volumes.

Definition. $\sqrt[n]{x}$ is the value $y$ with $y^{n} = x$ .

Square root ( $n = 2$ ): $\sqrt{x}$ .

Edexcel uses POSITIVE root only when written as $\sqrt{}$ .
$\sqrt{16} = 4$ (not $\pm 4$ ).
$\sqrt{x}$ requires $x \ge 0$ (no real square root of negative).

Cube root ( $n = 3$ ): $\sqrt[3]{x}$ .

Works for ALL real $x$ (positive, negative, zero).
$\sqrt[3]{27} = 3$ , $\sqrt[3]{-8} = -2$ .

Higher roots. Same idea. $\sqrt[n]{x^{n}} = x$ for $x \ge 0$ .

Estimating square roots (non-calculator paper).

For $\sqrt{50}$ :

$7^{2} = 49 < 50 < 64 = 8^{2}$ .
$\sqrt{50}$ between $7$ and $8$ , much closer to $7$ .
$7.1^{2} = 50.41$ → $\sqrt{50} \approx 7.07$ (≈ $7.1$ to 1 d.p.).

Trap the number between neighbouring perfect squares to estimate its root.

Memorise common roots.

Square roots	Cube roots
$\sqrt{4} = 2$	$\sqrt[3]{8} = 2$
$\sqrt{9} = 3$	$\sqrt[3]{27} = 3$
$\sqrt{16} = 4$	$\sqrt[3]{64} = 4$
$\sqrt{25} = 5$	$\sqrt[3]{125} = 5$
$\sqrt{36} = 6$	$\sqrt[3]{216} = 6$
$\sqrt{49} = 7$	$\sqrt[3]{343} = 7$
$\sqrt{64} = 8$	$\sqrt[3]{512} = 8$
$\sqrt{81} = 9$	$\sqrt[3]{729} = 9$
$\sqrt{100} = 10$	$\sqrt[3]{1000} = 10$
$\sqrt{121} = 11$
$\sqrt{144} = 12$
$\sqrt{169} = 13$
$\sqrt{196} = 14$
$\sqrt{225} = 15$

Worked qualitative. Why is $\sqrt[3]{-8} = -2$ but $\sqrt{-8}$ undefined?

For odd roots, $(-2)^{3} = -8$ . So $\sqrt[3]{-8} = -2$ is fine.
For even roots, $y^{2} \ge 0$ always (squares of reals are non-negative). So no real $y$ has $y^{2} = -8$ .
Even roots of negatives don't exist in the real numbers (only in the complex numbers, beyond IGCSE).

Edexcel tip. Square roots in answers stay as surds when not perfect (covered in surds topic). Don't approximate $\sqrt{17}$ as $4.12$ unless asked — keep it as $\sqrt{17}$ .

$\sqrt{x} \ge 0$ in 4MA1.
$\sqrt[3]{}$ works for negatives.
Even roots: $x \ge 0$ required.
Memorise to $15^{2}$ and $10^{3}$ .

How it’s examined

Index laws appear on EVERY 4MA1 paper. Standalone questions (3-5 marks) on simplifying expressions; embedded use throughout algebra (Topic 2), surds (later in Topic 1), and standard form. Examiner reports flag (1) confusing $-$ with reciprocal for negative indices, (2) computing $x^{0} = 0$ , and (3) computing fractional powers using power-first instead of root-first on Paper 1H.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H May/Jun 2023, 4MA1/1H Jan 2024. Last reviewed 2026-05-07.

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Step-by-step worked examples — Powers and Roots

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

1Apply index laws to simplify
Foundation• index laws
Question
Simplify $\dfrac{x^{8} \times x^{3}}{x^{5}}$ .
Step-by-step solution
1. Step 1
  $x^{8} \times x^{3} = x^{8+3} = x^{11}$ (multiplication: ADD indices).
2. Step 2
  $\dfrac{x^{11}}{x^{5}} = x^{11-5} = x^{6}$ (division: SUBTRACT indices).
Answer
$x^{6}$
Examiner tip
Edexcel mark schemes are strict: ADD for multiplication, SUBTRACT for division. Mistakes here cascade through algebraic questions.
2Evaluate a negative index
Higher• Adapted from 4MA1/1H May/Jun 2023 Q9• negative index
Question
Evaluate $\left(\dfrac{2}{3}\right)^{-2}$ .
Step-by-step solution
1. Step 1
  Negative index = reciprocal: $\left(\dfrac{a}{b}\right)^{-n} = \left(\dfrac{b}{a}\right)^{n}$ .
2. Step 2
  $\left(\dfrac{2}{3}\right)^{-2} = \left(\dfrac{3}{2}\right)^{2} = \dfrac{9}{4}$ .
Answer
$\dfrac{9}{4}$ or $2\dfrac{1}{4}$
Examiner tip
$x^{-n} = \dfrac{1}{x^{n}}$ . For fractions: flip first, then power. Mark scheme awards M1 for the reciprocal step, A1 for the answer.
3Evaluate a fractional index
Higher• fractional index, roots
Question
Evaluate $27^{2/3}$ .
Step-by-step solution
1. Step 1
  $x^{m/n} = \sqrt[n]{x^{m}} = (\sqrt[n]{x})^{m}$ .
2. Step 2
  Take the cube root first (smaller numbers): $27^{1/3} = 3$ .
3. Step 3
  Then square: $3^{2} = 9$ .
Answer
$9$
Examiner tip
ALWAYS take the root first, then power. Smaller numbers → less arithmetic error. $27^{2/3} = (27^{1/3})^{2} = 3^{2} = 9$ .
4Anything to the power of zero
Foundation• zero index
Question
Evaluate $5^{0}$ and $(-7)^{0}$ .
Step-by-step solution
1. Step 1
  $x^{0} = 1$ for any $x \ne 0$ .
2. Step 2
  Both equal $1$ — including the negative example.
Answer
$5^{0} = 1$ , $(-7)^{0} = 1$
Examiner tip
$x^{0} = 1$ is a definition that comes from the rule $x^{a-a} = \dfrac{x^{a}}{x^{a}} = 1$ . Don't write $x^{0} = x$ — that's the most common error.
5Combine multiple index laws
Higher• Adapted from 4MA1/1H Jan 2024 Q11• mixed indices
Question
Simplify $\dfrac{(2x^{3})^{2} \times x^{-1}}{x^{4}}$ .
Step-by-step solution
1. Step 1
  Expand the bracket: $(2x^{3})^{2} = 2^{2} \times x^{6} = 4x^{6}$ .
2. Step 2
  Substitute: $\dfrac{4x^{6} \times x^{-1}}{x^{4}}$ .
3. Step 3
  Numerator: $x^{6} \times x^{-1} = x^{5}$ . So $\dfrac{4x^{5}}{x^{4}}$ .
4. Step 4
  Divide: $\dfrac{4x^{5}}{x^{4}} = 4x^{5-4} = 4x$ .
Answer
$4x$
Examiner tip
Multi-step index questions: handle ONE rule at a time. Take care of the bracket first ( $(ab)^{n} = a^{n}b^{n}$ ), then combine in the numerator, then divide.
6Estimate a square root without a calculator
Foundation• square root, estimation
Question
Estimate $\sqrt{50}$ to 1 decimal place WITHOUT using a calculator. Justify your estimate.
Step-by-step solution
1. Step 1
  Find the nearest perfect squares: $7^{2} = 49$ and $8^{2} = 64$ .
2. Step 2
  $\sqrt{50}$ is between $7$ and $8$ , closer to $7$ .
3. Step 3
  Estimate: $\sqrt{50} \approx 7.1$ (since $7.1^{2} = 50.41$ , very close).
Answer
$\sqrt{50} \approx 7.1$
Examiner tip
Paper 1H is non-calculator. For square roots of non-perfect squares, find the bracket of perfect squares first, then estimate the position.

Key Formulae — Powers and Roots

The formulae you need to memorise for powers and roots on the Pearson Edexcel IGCSE 4MA1 paper, with every variable defined in plain English and a note on when to use it.

Product rule
$x^{a} \times x^{b} = x^{a+b}$
$x$
any non-zero base
$a, b$
any indices (positive, negative, fractional)
When to use
Multiplying powers of the SAME base. Add the indices.
Example
$x^{3} \times x^{5} = x^{8}$ .
Quotient rule
$\dfrac{x^{a}}{x^{b}} = x^{a-b}$
$x$
any non-zero base
When to use
Dividing powers of the SAME base. Subtract the indices.
Example
$\dfrac{x^{7}}{x^{2}} = x^{5}$ .
Power rule
$(x^{a})^{b} = x^{ab}$
$x$
any non-zero base
When to use
Raising a power to another power. Multiply the indices.
Example
$(x^{3})^{4} = x^{12}$ .
Zero index
$x^{0} = 1\ \text{(for}\ x \ne 0\text{)}$
$x$
any non-zero number
When to use
Always. Anything to the power $0$ is $1$ (except $0$ itself, which is undefined).
Example
$5^{0} = 1$ , $(-7)^{0} = 1$ , $(\pi)^{0} = 1$ .
Negative index
$x^{-n} = \dfrac{1}{x^{n}}$
$x$
any non-zero base
When to use
When you see a negative power. Take reciprocal, then apply positive power.
Example
$2^{-3} = \dfrac{1}{2^{3}} = \dfrac{1}{8}$ . $\left(\dfrac{2}{3}\right)^{-2} = \left(\dfrac{3}{2}\right)^{2} = \dfrac{9}{4}$ .
Fractional index
$x^{m/n} = \sqrt[n]{x^{m}} = (\sqrt[n]{x})^{m}$
$x$
non-negative if $n$ even
$m, n$
integers, $n > 0$
When to use
Whenever the index is a fraction. Take the root first (smaller numbers, fewer errors), then apply the power.
Example
$8^{2/3} = (8^{1/3})^{2} = 2^{2} = 4$ . $16^{3/4} = (16^{1/4})^{3} = 2^{3} = 8$ .
Product inside bracket
$(xy)^{n} = x^{n} \times y^{n}$
$x, y$
any non-zero bases
When to use
Distributing a power across a product (NOT a sum).
Example
$(2x)^{3} = 2^{3} \times x^{3} = 8x^{3}$ . NOTE: $(x+y)^{n} \ne x^{n} + y^{n}$ in general.

Key Definitions and Keywords — Powers and Roots

Definitions to memorise and the exact keywords mark schemes credit for powers and roots answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IGCSE 4MA1 sitting.

Base
In $x^{n}$ , the value being raised to a power.
Example
In $2^{5}$ , the base is $2$ .
Index (or exponent / power)
Examiner keyword
In $x^{n}$ , the value $n$ telling you how many times to multiply the base.
Example
In $2^{5}$ , the index is $5$ .
Square root
Examiner keyword
$\sqrt{x}$ — a value $y$ such that $y^{2} = x$ . Edexcel uses positive root only.
Example
$\sqrt{49} = 7$ (not $\pm 7$ in this notation).
Cube root
Examiner keyword
$\sqrt[3]{x}$ — a value $y$ such that $y^{3} = x$ .
Example
$\sqrt[3]{27} = 3$ . $\sqrt[3]{-8} = -2$ (cube root of negative is real).
Perfect square
An integer that is the square of another integer.
Example
$1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, \ldots$

Common Mistakes and Misconceptions — Powers and Roots

The traps other students keep falling into on powers and roots questions — taken from recent Pearson Edexcel IGCSE 4MA1 examiner reports and mark schemes — and how to avoid them.

✕Adding indices when multiplying different bases
4MA1/1H — examiner reports 2022-2024
Why it happens
Students apply $x^{a} \times x^{b} = x^{a+b}$ to different bases too.
How to avoid it
Index laws ONLY work for the SAME base. $x^{3} \times y^{4}$ doesn't simplify; $x^{3} \times x^{4} = x^{7}$ .
✕Writing $x^{0} = 0$ or $x^{0} = x$
Why it happens
Zero suggests 'nothing'; some students think $0$ powers cancel.
How to avoid it
Memorise: $x^{0} = 1$ for any non-zero $x$ . Reason: $\dfrac{x^{n}}{x^{n}} = x^{n-n} = x^{0}$ , and any number divided by itself is $1$ .
✕Writing $x^{-n} = -x^{n}$
4MA1/1H May/Jun 2023 — examiner report Q9
Why it happens
Students confuse 'negative index' with 'negative number'.
How to avoid it
Negative index means RECIPROCAL: $x^{-n} = \dfrac{1}{x^{n}}$ . The base stays positive (if it was positive).
✕Computing $27^{2/3}$ as $(27^{2})^{1/3}$ on a non-calculator paper
4MA1/1H — examiner reports 2022-2024
Why it happens
Both methods give the same answer, but $27^{2} = 729$ is harder to take cube root of mentally.
How to avoid it
ALWAYS take the root FIRST: $27^{1/3} = 3$ , then square to get $9$ . Smaller numbers, less error.
✕Forgetting to apply the power to constants in brackets
4MA1/1H Jan 2024 — examiner report Q11
Why it happens
Students apply the power only to the variable: $(2x)^{3} = 2x^{3}$ (wrong).
How to avoid it
$(ab)^{n} = a^{n} \cdot b^{n}$ . So $(2x)^{3} = 2^{3} \cdot x^{3} = 8x^{3}$ . The power applies to EVERYTHING in the bracket.
✕Writing $(x+y)^{2} = x^{2} + y^{2}$
Why it happens
Confusing with the product rule.
How to avoid it
Index laws DON'T distribute over addition. $(x+y)^{2} = x^{2} + 2xy + y^{2}$ (use FOIL).

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Powers and Roots

Short Study Notes in the form of Slides

Short Study Notes — Powers and Roots

Detailed Notes

What you’ll learn

How it’s examined

1Apply index laws to simplify

2Evaluate a negative index

3Evaluate a fractional index

4Anything to the power of zero

5Combine multiple index laws

6Estimate a square root without a calculator

Product rule

Quotient rule

Power rule

Zero index

Negative index

Fractional index

Product inside bracket

Base

Index (or exponent / power)

Square root

Cube root

Perfect square

✕Adding indices when multiplying different bases

✕Writing x0=0x^{0} = 0x0=0 or x0=xx^{0} = xx0=x

✕Writing x−n=−xnx^{-n} = -x^{n}x−n=−xn

✕Computing 272/327^{2/3}272/3 as (272)1/3(27^{2})^{1/3}(272)1/3 on a non-calculator paper

✕Forgetting to apply the power to constants in brackets

✕Writing (x+y)2=x2+y2(x+y)^{2} = x^{2} + y^{2}(x+y)2=x2+y2

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Powers and Roots — frequently asked questions

✕Writing $x^{0} = 0$ or $x^{0} = x$

✕Writing $x^{-n} = -x^{n}$

✕Computing $27^{2/3}$ as $(27^{2})^{1/3}$ on a non-calculator paper

✕Writing $(x+y)^{2} = x^{2} + y^{2}$