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# Squares, cubes, square roots, and cube roots of algebraic terms

## 8.3 Squares, cubes, square roots, and cube roots of algebraic terms

Remember that when we take the square root of a number, we are looking for a number that, when multiplied by itself, gives the number within the square root. For example, $$\sqrt{9} = 3$$ because $$3 \times 3 = 9$$.

When we take the cube root of a number, we are looking for a number that, when multiplied by itself, and by itself again, gives the number within the cube root. For example, $$\sqrt{27} = 3$$ because $$3 \times 3 \times 3 = 27$$.

## Worked Example 8.9: Finding square roots of algebraic terms

Simplify the following expression as much as possible:

$\sqrt{9y^{10}}$

### Expand the power and group the terms

We will expand the power and try to group the bases into two equal groups.

We must write the constant coefficient as the product of its prime factors.

\begin{align} & \sqrt{9y^{10}} \\ = &\ \sqrt{3 \times 3 \times y \times y \times y \times y \times y \times y \times y \times y \times y \times y} \\ = &\ \sqrt{(3 \times y \times y \times y \times y \times y) \times (3 \times y \times y \times y \times y \times y)} \\ = &\ \sqrt{\left( 3y^{5} \right) \times \left( 3y^{5} \right)} \\ = &\ \sqrt{\left( 3y^{5} \right)^{2}} \\ = &\ 3y^{5} \end{align}

## Worked Example 8.10: Finding cube roots of algebraic terms

Simplify the following expression:

$\sqrt{t^{6}}$

### Expand the power and group the bases

Expand the power, then group the bases into three equal groups.

We will first expand the power. Then, we will try to group the bases into three equal groups.

\begin{align} & \sqrt{t^{6}} \\ = &\ \sqrt{t \times t \times t \times t \times t \times t} \\ = &\ \sqrt{(t \times t) \times (t \times t) \times (t \times t)} \\ = &\ \sqrt{t^{2} \times t^{2} \times t^{2}} \\ = &\ t^{2} \end{align}

Check that your answer makes sense. If we multiply our answer by itself three times, we should get what we started with inside the cube root. We have $$t^{2} \times t^{2} \times t^{2} = t^{6}$$ so our answer makes sense!

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Sometimes you will need to perform operations within the root, such as addition, subtraction, multiplication, or division of algebraic terms. You should do these operations first before taking the square or cube root.

## Worked Example 8.11: Finding roots of algebraic fractions

Simplify the following expression as much as possible:

$\sqrt{\frac{125h^{9}}{8}}$

### Group the bases into three equal groups.

\begin{align} & \sqrt{\frac{125h^{9}}{8}} \\ &= \sqrt{\frac{5h^{3}}{2} \times \frac{5h^{3}}{2} \times \frac{5h^{3}}{2}} \end{align}

### Give the cube root of the expression.

\begin{align} &= \sqrt{\left( \frac{5h^{3}}{2} \right)^{3}} \\ &= \frac{5h^{3}}{2} \end{align}

## Worked Example 8.12: Finding roots of a product

Simplify the following expression:

$\sqrt{3t^{2} \times 9t^{4}}$

### Simplify the expression within the root first.

\begin{align} & \sqrt{3t^{2} \times 9t^{4}} \\ &=\ \sqrt{27t^{6}} \end{align}

### Expand the power and try to group the bases into three equal groups.

We must write the constant coefficient as the product of its prime factors.

\begin{align} & \sqrt{3t^{2} \times 9t^{4}} \\ &= \sqrt{27t^{6}} \\ &= \sqrt{3 \times 3 \times 3 \times t \times t \times t \times t \times t \times t} \\ &= \sqrt{(3 \times t \times t) \times (3 \times t \times t) \times (3 \times t \times t)} \\ &= \sqrt{\left( 3t^{2} \right) \times \left( 3t^{2} \right) \times \left( 3t^{2} \right)} \\ &= 3t^{2} \end{align}

## Worked Example 8.13 Finding roots of a sum

Simplify the following expression:

$\sqrt{8t^{6} + 19t^{6}}$

### Add the terms within the cube root.

The terms within the root are like terms, so we will first add them together.

\begin{align} & \sqrt{8t^{6} + 19t^{6}} \\ &= \sqrt{27t^{6}} \end{align}

### Expand the power and try to group the bases into three equal groups.

\begin{align} & \sqrt{8t^{6} + 19t^{6}} \\ &= \sqrt{27t^{6}} \\ &= \sqrt{3 \times 3 \times 3 \times t \times t \times t \times t \times t \times t} \\ &= \sqrt{(3 \times t \times t) \times (3 \times t \times t) \times (3 \times t \times t)} \\ &= \sqrt{3t^{2} \times 3t^{2} \times 3t^{2}} \\ &= 3t^{2} \end{align}

## Worked Example 8.14: Finding roots of a quotient

Simplify the following expression:

$\sqrt{\frac{t^{7}}{t}}$

Then evaluate the expression for $$t = 4$$.

### First divide the powers within the root.

We can simplify the term within the root by dividing the powers of the same base.

\begin{align} & \sqrt{\frac{t^{7}}{t}} \\ = & \sqrt{t^{6}} \end{align}

### Expand the simplified power and try to group the bases into two equal groups.

\begin{align} &= \sqrt{t^{6}} \\ &= \sqrt{t \times t \times t \times t \times t \times t} \\ &= \sqrt{\left( t^{3} \right) \times \left( t^{3} \right)} \\ &= t^{3} \end{align}

### Evaluate.

Use the simplified expression from Step 2.

In Step 2 we determined that $$\sqrt{\frac{t^{7}}{t}}$$ is equal to $$t^{3}$$. Therefore, we can substitute $$t = 4$$ into the simplified expression:

\begin{align} \sqrt{\frac{t^{7}}{t}} &= t^{3} \\ &= (4)^{3} \\ &= 64 \end{align}

You could also substitute $$t = 4$$ into the original expression. Your answer would be the same, but your calculations would be longer.

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