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Cambridge IGCSE Mathematics — 0580 Extended

Topic 1.3: Number — Powers & Roots

Squares and square roots

\(n^2\) means \(n \times n\). \(\sqrt{n}\) is the non-negative value that squares to \(n\) (for \(n \geq 0\)).

\(n\)\(n^2\)\(n\)\(n^2\)
11981
2410100
3911121
41612144
52513169
63614196
74915225
864

Paper 2 (non-calculator)

Knowing \(13^2 = 169\) and \(\sqrt{144} = 12\) saves time — drill the table until recall is instant.

Cubes and cube roots

\(n\)\(n^3\)\(\sqrt[3]{n^3}\)
111
282
3273
4644
51255
10100010

Calculating with powers

Method

  1. Evaluate roots and powers inside brackets first (respect BIDMAS).
  2. Combine same-base powers using index laws.
  3. Write the final answer as an integer or simplified power.

Evaluate \(2^3 \times 2^4\).

Two cubed times two to the fourth equals two to the seventh which is one hundred twenty eight
Same base — add the indices, then evaluate 2⁷ = 128.

Evaluate \(\sqrt{49} + 2^3 \times \sqrt[3]{27}\).

Square root of forty nine plus two cubed times cube root of twenty seven equals thirty one with BIDMAS order
Evaluate roots and powers first, then multiply before adding — 7 + 8 × 3 = 31.

Try this

Simplify \(\dfrac{5^6}{5^2}\).

Show answer
Answer

  1. Same base — subtract the indices

    \[ \dfrac{5^{6}}{5^{2}} = 5^{6-2} = 5^{4} \]

  2. Evaluate the power

    \[ 5^{4} = 5 \times 5 \times 5 \times 5 = 625 \]

  3. Answer

    \[ 625 \]

Exam Traps

  • \(2^3 \times 3^2 \neq 6^5\) — index laws apply only when the base is the same.

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