NZ Level 5
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Simplify numerical square and cube root expressions

The square root function reverses the squaring function. Similarly, the cube root function undoes the cubing function.

In algebraic notation, we write $\sqrt{a^2}=a$a2=a and $\sqrt[3]{b^3}=b$3b3=b. It is also true that $\left(\sqrt{a}\right)^2=a$(a)2=a and $\left(\sqrt[3]{b}\right)^3=b$(3b)3=b.

Square root and cube root expressions can sometimes be written in simpler forms using these facts, together with the other familiar surd rules.

We can use these ideas to simplify expressions like $\sqrt{16}$16 and $\sqrt{121}$121, where the numbers in the surd are perfect squares, or expressions like $\sqrt[3]{27}$327 and $\sqrt[3]{125}$3125, where the numbers in the surd are perfect cubes.

But what if the number in the surd contains a factor that is a perfect square or perfect cube? For example, an expression like $\sqrt{32}=\sqrt{16\times2}$32=16×2. Can we still simplify these types of expressions?



One method to check whether an expression can be simplified is by looking at its prime factors. For example, given the expression $\sqrt{18}$18, we first look at the prime factors of $18$18 which gives us $3\times3\times2=3^2\times2$3×3×2=32×2.

Now we can write $\sqrt{18}$18 as $\sqrt{3^2\times2}=\sqrt{3^2}\times\sqrt{2}$32×2=32×2. In this step we have used the important fact that $\sqrt{a\times b}=\sqrt{a}\times\sqrt{b}$a×b=a×b.

The first term in the product $\sqrt{3^2}\times\sqrt{2}$32×2 is of the form $\sqrt{a^2}=a$a2=a, so the fully simplified expression becomes $3\sqrt{2}$32.

From this example we can see that if any factor appears two times within a square root, or three times within a cube root, then the expression can be further simplified. This is equivalent to looking for a factor that is a perfect square for square root expressions, or a perfect cube for cube root expressions.


Worked example 1

Simplify $\sqrt{8}$8.

Think: If we can find any factors of $8$8 that are perfect squares, then we can simplify the expression using the fact that $\sqrt{a^2b}=a\sqrt{b}$a2b=ab.

Do: The factors of $8$8 are $1$1, $2$2, $4$4, and $8$8. Let's use the perfect square $4$4 to rewrite the expression and simplify.

$\sqrt{8}$8 $=$= $\sqrt{4\times2}$4×2 (Replace $8$8 with two factors)
  $=$= $\sqrt{4}\times\sqrt{2}$4×2 (Use the fact that $\sqrt{a\times b}=\sqrt{a}\times\sqrt{b}$a×b=a×b)
  $=$= $\sqrt{2^2}\times\sqrt{2}$22×2 (Rewrite $4$4 as $2^2$22)
  $=$= $2\sqrt{2}$22 (Use the fact that $\sqrt{a^2}=a$a2=a)


Worked example 2

Simplify $\sqrt{392}$392.

Think: Looking at the prime factors gives us $392=2^3\times7^2$392=23×72. Notice that we can write $2^3$23 as $2\times2^2$2×22.

Do: After we substitute $392$392 with its prime factors we can use the rules of manipulation of surds to simplify the expression.

$\sqrt{392}$392 $=$= $\sqrt{2\times2^2\times7^2}$2×22×72
  $=$= $2\times7\times\sqrt{2}$2×7×2
  $=$= $14\sqrt{2}$142
Rules for manipulation of surds

Here are some key identities that we will find useful to simplify expressions involving surds.


$\sqrt{a\times b}=\sqrt{a}\times\sqrt{b}$a×b=a×b

And from these two we can see that $\sqrt{a^2b}=a\sqrt{b}$a2b=ab.


Practice questions

Question 1

Simplify $\sqrt{25}$25.

question 2

Simplify $\sqrt[3]{8\times6}$38×6.

Question 3

Simplify $\frac{64}{\sqrt{64}}$6464.



Use prime numbers, common factors and multiples, and powers (including square roots)

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