Powers, Roots and Index Laws

This lesson covers squares, cubes, square roots, cube roots, and the index laws (including negative, zero and fractional indices for Higher tier). These topics are essential for the Edexcel GCSE Mathematics (1MA1) specification and are frequently tested across all three papers.

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Key Vocabulary

Term	Meaning
Power / Index / Exponent	The small number written above and to the right, e.g. the 3 in $2^{3}$23
Base	The number being raised to a power, e.g. the 2 in $2^{3}$23
Square number	A number multiplied by itself, e.g. $5^{2} = 25$52=25
Cube number	A number multiplied by itself three times, e.g. $4^{3} = 64$43=64
Square root ($\sqrt{\phantom{x}}$x​)	The inverse of squaring: $\sqrt{25} = 5$25​=5
Cube root ($\sqrt[3]{\phantom{x}}$3x​)	The inverse of cubing: $\sqrt[3]{64} = 4$364​=4

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Square Numbers and Square Roots

Square Numbers to Know

n	$n^{2}$n2
1	1
2	4
3	9
4	16
5	25
6	36
7	49
8	64
9	81
10	100
11	121
12	144
13	169
14	196
15	225

Edexcel Exam Tip: You MUST know all square numbers up to $15^{2} = 225$152=225 for Paper 1 (non-calculator). These appear constantly in questions on surds, Pythagoras' theorem, and algebra.

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Cube Numbers and Cube Roots

Cube Numbers to Know

n	$n^{3}$n3
1	1
2	8
3	27
4	64
5	125
10	1000

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Index Laws

The index laws are rules for simplifying expressions involving powers. In these laws, a and b are non-zero numbers and m, n are integers (or rational numbers for Higher).

Law 1: Multiplication (Same Base)

$a^m \times a^n = a^{m+n}$am×an=am+n

When multiplying powers with the same base, ADD the indices.

Worked Example 1

Simplify $3^{4} \times 3^{2}$34×32

$3^{4} \times 3^{2} = 3^{4+2} = 3^{6} = 729$34×32=34+2=36=729

Law 2: Division (Same Base)

$a^m \div a^n = a^{m-n}$am÷an=am−n

When dividing powers with the same base, SUBTRACT the indices.

Worked Example 2

Simplify $5^{7} \div 5^{3}$57÷53

$5^{7} \div 5^{3} = 5^{7-3} = 5^{4} = 625$57÷53=57−3=54=625

Law 3: Power of a Power

$(a^m)^n = a^{m \times n}$(am)n=am×n

When raising a power to another power, MULTIPLY the indices.

Worked Example 3

Simplify $(2^{3})^{4}$(23)4

$(2^{3})^{4} = 2^{3\times4} = 2^{12} = 4096$(23)4=23×4=212=4096

Law 4: Power of a Product

$(ab)^n = a^n \times b^n$(ab)n=an×bn

Worked Example 4

Simplify $(3x)^{2}$(3x)2

$(3x)^{2} = 3^{2} \times x^{2} = 9x^{2}$(3x)2=32×x2=9x2

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The Zero Index

$a^{0} = 1$a0=1 (for any non-zero a)

Why?

Using the division law: $a^{3} \div a^{3} = a^{3-3} = a^{0}$a3÷a3=a3−3=a0

But $a^{3} \div a^{3} = 1 ($a3÷a3=1(any number divided by itself is 1)

Therefore $a^{0} = 1$a0=1.

Worked Example 5

• $7^{0} = 1$70=1
• $(0.5)^{0} = 1$(0.5)0=1
• $100^{0} = 1$1000=1

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Negative Indices [H]

a^{-n} = 1/a^n

A negative index means "the reciprocal".

Why?

$a^{2} \div a^{5} = a^{2-5} = a^{-3}$a2÷a5=a2−5=a−3

But $a^{2} \div a^{5} = a^{2}/a^{5} = 1/a^{3}$a2÷a5=a2/a5=1/a3

Therefore $a^{-3} = 1/a^{3}$a−3=1/a3.

Worked Example 6 [H]

Evaluate 2^{-3}

$2^{-3} = 1/2^{3} = 1/8$2−3=1/23=1/8

Worked Example 7 [H]

Evaluate 5^{-2}

$5^{-2} = 1/5^{2} = 1/25$5−2=1/52=1/25

Worked Example 8 [H]

Simplify (3/4)^{-2}

$(3/4)^{-2} = (4/3)^{2} = 16/9$(3/4)−2=(4/3)2=16/9

Key Rule: A fraction to a negative power means "flip the fraction and make the power positive."

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Fractional Indices [H]

$a^{1/n} = \sqrt[n]{a}$a1/n=na​ (the nth root of a)

$a^{m/n} = (\sqrt[n]{a})^m = \sqrt[n]{a^m}$am/n=(na​)m=nam​

Worked Example 9 [H]

Evaluate 27^{1/3}

$27^{1/3} = \sqrt[3]{27} = 3$271/3=327​=3

Worked Example 10 [H]

Evaluate 16^{3/4}

Method: Find the root first (usually easier), then raise to the power.

$16^{3/4} = (\sqrt[4]{16})^{3} = 2^{3} = 8$163/4=(416​)3=23=8

Worked Example 11 [H]

Evaluate 8^{-2/3}

Step 1: Deal with the fraction: $8^{2/3} = (\sqrt[3]{8})^{2} = 2^{2} = 4$82/3=(38​)2=22=4

Step 2: Deal with the negative: 8^{-2/3} = 1/4

Answer: 1/4

Edexcel Exam Tip: For fractional indices, always do the ROOT first, then the power. This keeps the numbers smaller and easier to work with. A very common Higher-tier question is evaluating expressions like 32^{3/5} or 27^{-2/3}.

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Combining Index Laws

Worked Example 12

Simplify $(2x^{3}y)^{2} \times 3x^{2}y^{4}$(2x3y)2×3x2y4

Step 1: Expand the bracket: $(2x^{3}y)^{2} = 4x^{6}y^{2}$(2x3y)2=4x6y2

Step 2: Multiply: $4x^{6}y^{2} \times 3x^{2}y^{4} = 12x^{8}y^{6}$4x6y2×3x2y4=12x8y6

Answer: $12x^{8}y^{6}$12x8y6

Worked Example 13 [H]

Simplify $(8a^{6})^{2/3}$(8a6)2/3

$= 8^{2/3} \times a^{6 \times 2/3}$=82/3×a6×2/3

$= (\sqrt[3]{8})^{2} \times a^{4}$=(38​)2×a4

$= 4a^{4}$=4a4

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Common Mistakes and Misconceptions

Mistake	Correction
$2^{3} \times 2^{4} = 2^{12}$23×24=212	ADD indices when multiplying same base: $2^{3} \times 2^{4} = 2^{7}$23×24=27
$3^{2} \times 2^{3} = 6^{5}$32×23=65	Cannot use index laws with DIFFERENT bases
$(-3)^{2} = -9$(−3)2=−9	$(-3)^{2} = (-3) \times (-3) = 9$(−3)2=(−3)×(−3)=9
16^{1/2} = 8	$16^{1/2} = \sqrt{16} = 4 ($161/2=16​=4(square root, not "half of")
2^{-3} = -8	2^{-3} = 1/8 (reciprocal, not negative)
$(2 + 3)^{2} = 4 + 9 = 13$(2+3)2=4+9=13	$(2 + 3)^{2} = 5^{2} = 25 ($(2+3)2=52=25(cannot "distribute" a power over addition)

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Practice Problems

1. Evaluate: $2^{5} \times 2^{3}$25×23
2. Simplify: $x^{7} \div x^{2}$x7÷x2
3. Evaluate: $10^{0}$100
4. [H] Evaluate: 4^{-3}
5. [H] Evaluate: 125^{2/3}
6. [H] Evaluate: (9/16)^{-1/2}
7. Simplify: $(3a^{2}b)^{3}$(3a2b)3
8. [H] Simplify: $(27x^{9})^{1/3}$(27x9)1/3

Answers

1. $2^{8} = 256$28=256
2. $x^{5}$x5
3. 1
4. 1/64
5. $(\sqrt[3]{125})^{2} = 5^{2} = 25$(3125​)2=52=25
6. (16/9)^{1/2} = 4/3
7. $27a^{6}b^{3}$27a6b3
8. $3x^{3}$3x3

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Summary