POP

E1.3 Powers and Roots

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01 — Explanation

What squares and cubes do I need to know from memory?

1^2,2^2,\dots,15^2 = 1,4,9,16,25,36,49,64,81,100,121,144,169,196,225

1^3,2^3,3^3,4^3,5^3,10^3 = 1,8,27,64,125,1000

Instant recall of these saves precious time on non-calculator papers and underpins index laws, surds, and standard-form questions throughout the syllabus.

02 — Explanation

How is \sqrt{a} different from the solutions to x^2 = a?

The symbol $\sqrt{a}$ always means the positive square root only. But the equation $x^2 = a$ has two solutions, $x = \pm\sqrt{a}$ , because both a positive and a negative number square to give $a$ .

Examiner Tip: Only write $\pm$ when you are solving an equation. If a question just asks you to evaluate $\sqrt{36}$ , the answer is $6$ , not $\pm6$ .

Cube roots behave differently — $\sqrt[3]{a}$ is defined for negative $a$ too, since a negative number cubed stays negative: $\sqrt[3]{-27} = -3$ .

03 — Explanation

How do I evaluate an expression with mixed roots and powers?

Work out each root or power separately first, respecting the normal order of operations, then combine.

Worked Example

Work out the value of

\frac{\sqrt{169} + \sqrt[3]{125}}{2^3 - \sqrt{36}}

Evaluate each part from recall: $\sqrt{169}=13$ , $\sqrt[3]{125}=5$ , $\sqrt{36}=6$ .
Numerator: $13+5=18$ .
Denominator: $2^3-6 = 8-6=2$ .
Divide: $\dfrac{18}{2}=9$ .

Examiner Tip: A root symbol only applies to what's directly underneath it. Don't be tempted to lump $2^3 - \sqrt{36}$ together as $\sqrt{2^3 - 36}$ — evaluate the power and the root independently before subtracting.

04 — Explanation

How do I estimate the value of a root that isn't a perfect square, without a calculator?

Find the two consecutive perfect squares (or cubes) that the number sits between, and use them to trap the root between two consecutive integers.

Worked Example

Without using a calculator, show that $\sqrt{130}$ lies between $11$ and $12$ .

Find nearby perfect squares: $11^2 = 121$ and $12^2 = 144$ .
Since $121 < 130 < 144$ , taking square roots of all three parts preserves the inequality: $\sqrt{121} < \sqrt{130} < \sqrt{144}$ .
So $11 < \sqrt{130} < 12$ .

You can sharpen the estimate further by noticing $130$ is much closer to $121$ than to $144$ , so $\sqrt{130} \approx 11.4$ (the true value is $11.40\ldots$ ).

05 — Explanation

How do I handle powers and roots that are combined inside a single set of brackets?

Work out the contents of the bracket completely first, then apply the outer power or root to that single resulting value — never apply it to each term separately.

Worked Example

Evaluate $(\sqrt{25} - \sqrt[3]{8})^3$ .

Evaluate inside the brackets first: $\sqrt{25} = 5$ and $\sqrt[3]{8} = 2$ , so the bracket becomes $5 - 2 = 3$ .
Cube the single resulting value: $3^3 = 27$ .

Examiner Tip: Writing $(\sqrt{25})^3 - (\sqrt[3]{8})^3$ instead — cubing each term separately before subtracting — gives a completely different (and wrong) answer of $117$ . Brackets always mean "simplify inside first".

06 — Explanation

Why is it worth memorising squares and cubes rather than always using a calculator?

On non-calculator papers, instant recall is the only way to evaluate these efficiently, but even on calculator papers, recognising that $169$ is $13^2$ or that $64$ is both $8^2$ and $4^3$ lets you spot simplification opportunities — such as a hidden perfect square inside a surd, or a hidden perfect cube inside a cube root — far faster than typing every option into a calculator to check.