Lesson Explainer: Operations on Cube Roots Mathematics

In this explainer, we will learn how to use the properties of operations on cube roots to simplify expressions.

We begin by recalling that, for any two real numbers π‘Ž and 𝑏, we have that οŽ’οŽ’οŽ’βˆšπ‘ŽΓ—βˆšπ‘=βˆšπ‘ŽΓ—π‘. We can prove this fact by noting that ο€»βˆšπ‘ŽΓ—βˆšπ‘ο‡=βˆšπ‘ŽΓ—βˆšπ‘Γ—βˆšπ‘ŽΓ—βˆšπ‘Γ—βˆšπ‘ŽΓ—βˆšπ‘=ο€Ίβˆšπ‘Žο†Γ—ο€»βˆšπ‘ο‡=π‘ŽΓ—π‘.

We can write this formally as follows.

Property: The Product of Cube Roots

For any real numbers π‘Ž and 𝑏, we have that οŽ’οŽ’οŽ’βˆšπ‘ŽΓ—βˆšπ‘=βˆšπ‘ŽΓ—π‘.

We can use this result in both directions to simplify radical expressions.

First, we can simplify the product of radical expressions. Consider the following example: √12Γ—βˆš18=√12Γ—18=√216=6.

Second, we can simplify a cube root that is divisible by a perfect cube by splitting it into a product of cube roots. For example, √54=√27Γ—2=√27Γ—βˆš2=3√2.

By using the second method, we can rewrite the cube root of a non-perfect cube in a simplified form. Consider οŽ’βˆšπ‘ for an integer 𝑐; we can find the greatest perfect cube that divides 𝑐, for example, π‘ŽοŠ©. Then, we have that οŽ’οŽ’οŽ’βˆšπ‘=βˆšπ‘ŽΓ—π‘=π‘Žβˆšπ‘, where 𝑏=π‘π‘ŽοŠ©.

We want the sign of π‘Ž to represent the sign of the number, so we choose our perfect cube to make π‘Ž the same sign as 𝑐. This gives us the following method for simplifying the cube root of a non-perfect cube.

How To: Simplifying the Cube Root of a Non-Perfect Cube

Given any integer 𝑐, we write οŽ’βˆšπ‘ in the form π‘Žβˆšπ‘οŽ’, where 𝑏 is the smallest possible positive integer, by using the following steps:

  1. Find the greatest perfect cube that divides 𝑐.
  2. Choose integer π‘Ž to have the same sign as 𝑐 such that π‘ŽοŠ© divides 𝑐 and π‘ŽοŠ© is the greatest perfect cube dividing 𝑐.
  3. We then have that οŽ’οŽ’οŽ’βˆšπ‘=βˆšπ‘ŽΓ—π‘=π‘Žβˆšπ‘οŠ©.

It is worth noting that we do not immediately need to identify the greatest perfect cube dividing the radicand of the cube root (the number inside the cube root) straight away. This is because we can apply this process multiple times. For example, √1080=√8Γ—135=2√135=2√27Γ—5=2ο€»3√5=6√5.

Let’s now see an example of applying this process to simplify the cube root of integer values.

Example 1: Expressing a Cube Root in Its Simplest Form

Write each of the following radical expressions in the form π‘Žβˆšπ‘οŽ’, where π‘Ž and 𝑏 are integers and 𝑏 is the smallest possible positive value. √256βˆšβˆ’540

Answer

Part 1

We start by looking for perfect cubes that divide 256. We see that 2=8 is a factor of 256 since 256=8Γ—32, and we recall that, for any two real numbers 𝑛 and π‘š, we have that οŽ’οŽ’οŽ’βˆšπ‘›Γ—π‘š=βˆšπ‘›Γ—βˆšπ‘š. Hence, √256=√8Γ—32=√8Γ—βˆš32=2√32.

We might be tempted to stop here; however, we need 𝑏 to be the smallest possible positive value such that no perfect cubes can divide 𝑏. We see that 32=8Γ—4, so we can apply this process again. We have 2√32=2√8Γ—4=2ο€»βˆš8Γ—βˆš4=2ο€»2√4=4√4.

By inspection, we can see that there are no more perfect cubes that divide 4, so we cannot simplify further. Hence, √256=4√4.

Part 2

We want to apply the same process; however, this time, since the radicand of the cube root is negative, we need to find a negative perfect cube divisor of βˆ’540. We note that βˆ’540=(βˆ’27)Γ—20. Hence, οŽ’οŽ’οŽ’οŽ’οŽ’βˆšβˆ’540=√(βˆ’27)Γ—20=βˆšβˆ’27Γ—βˆš20=βˆ’3√20.

By inspection, we can see that there are no more perfect cubes that divide 20, so we cannot simplify further. Hence, οŽ’οŽ’βˆšβˆ’540=βˆ’3√20.

In our next example, we will apply the property of the product of cube roots to simplify a product of radicals.

Example 2: Simplifying the Product of Two Cube Roots

Express √4Γ—βˆšβˆ’16 in its simplest form.

Answer

We first recall that, for any two real numbers π‘Ž and 𝑏, we have that οŽ’οŽ’οŽ’βˆšπ‘ŽΓ—βˆšπ‘=βˆšπ‘ŽΓ—π‘. Applying this, we have √4Γ—βˆšβˆ’16=√4Γ—(βˆ’16)=βˆšβˆ’64.

We then note that (βˆ’4)=βˆ’64; hence, οŽ’βˆšβˆ’64=βˆ’4.

Thus far, we have worked with simplifying the cube roots of integers. However, this result works with the cube roots of any real numbers. An application of this result is to consider the cube root of the quotient of real numbers. We have οŽ’οŽ’οŽ’οŽ’ο„žπ‘Žπ‘=ο„žπ‘ŽΓ—ο€Ό1π‘οˆ=βˆšπ‘ŽΓ—ο„ž1𝑏.

We note that ο€Ώ1βˆšπ‘ο‹=ο€Ώ1βˆšπ‘ο‹Γ—ο€Ώ1βˆšπ‘ο‹Γ—ο€Ώ1βˆšπ‘ο‹=1π‘οŽ’οŽ’οŽ’οŽ’οŠ©. Hence, οŽ’οŽ’οŽ’οŽ’οŽ’οŽ’βˆšπ‘ŽΓ—ο„ž1𝑏=βˆšπ‘ŽΓ—ο€Ώ1βˆšπ‘ο‹=βˆšπ‘Žβˆšπ‘.

Of course, we assume that 𝑏≠0 since we are dividing by 𝑏. This gives us the following result.

Property: Cube Root of the Quotient of Real Numbers

For any real numbers π‘Ž and 𝑏, where 𝑏≠0, we have that οŽ’οŽ’οŽ’ο„žπ‘Žπ‘=βˆšπ‘Žβˆšπ‘.

Now, let’s look at an example of using both properties to simplify an expression involving radicals.

Example 3: Simplifying the Multiplication and Division of Multiple Cube Roots

Simplify √2Γ—βˆš4√32Γ—βˆšβˆ’2.

Answer

We start by noting that none of the radicands are perfect cubes, so we cannot directly evaluate any of the individual radicals. Instead, we will combine the radicals and then simplify. We first recall that, for any real numbers π‘Ž and 𝑏, we have that οŽ’οŽ’οŽ’βˆšπ‘ŽΓ—βˆšπ‘=βˆšπ‘ŽΓ—π‘. Applying this, we have √2Γ—βˆš4√32Γ—βˆšβˆ’2=√2Γ—4√32Γ—(βˆ’2)=√8βˆšβˆ’64.

We can simplify this expression by prime factoring the radicand of each cube root. We note that 8=2 and βˆ’64=(βˆ’4). So, √8βˆšβˆ’64=√2(βˆ’4)=2βˆ’4=βˆ’12.

In our next example, we will see how this process can be used to simplify the difference between the cube roots of integers.

Example 4: Simplifying the Difference of Two Cube Roots with Different Bases

Express √256βˆ’βˆš4 in its simplest form.

Answer

We first note that neither radicand of the cube roots is a perfect cube, so we cannot evaluate either term directly. Instead, let’s simplify each term by writing them in the form π‘Žβˆšπ‘οŽ’, where π‘Ž and 𝑏 are integers and 𝑏 is the smallest possible positive integer.

Now, we look for perfect cubes that divide 256; we find that 256=8Γ—32. We can then use the fact that, for any real numbers π‘Ž and 𝑏, we have that οŽ’οŽ’οŽ’βˆšπ‘ŽΓ—βˆšπ‘=βˆšπ‘ŽΓ—π‘ to simplify as follows: √256=√8Γ—32=√8Γ—βˆš32=2√32.

We can simplify this further by noting that 32=8Γ—4. Thus, 2√32=2√8Γ—4=2ο€»βˆš8Γ—βˆš4=2ο€»2√4=4√4.

Substituting this value into the given expression yields √256βˆ’βˆš4=4√4βˆ’βˆš4.

We then take out a factor of √4 to get 4√4βˆ’βˆš4=(4βˆ’1)√4=3√4.

We cannot find a perfect cube divisor of 4, which is greater than 1, so we cannot simplify the radical any further. Hence, √256βˆ’βˆš4=3√4.

In our next example, we will simplify an expression involving the addition of multiple radical expressions.

Example 5: Simplifying the Addition of Cube Roots

Express βˆ’5√192+5βˆšβˆ’648+√375 in its simplest form.

Answer

We will start by simplifying each term separately; we can do this by using the fact that, for any real numbers π‘Ž and 𝑏, we have that οŽ’οŽ’οŽ’βˆšπ‘ŽΓ—βˆšπ‘=βˆšπ‘ŽΓ—π‘.

To do this, we check 192 for any perfect cube divisors. We see that 8 is a factor of 192 since 192=8Γ—24. Hence, βˆ’5√192=βˆ’5ο€»βˆš8Γ—24=βˆ’5ο€»βˆš8Γ—βˆš24=βˆ’5ο€»2√24=βˆ’10√24.

We want the radicand of the cube root to be the smallest positive integer possible. We can do this by checking to see if 24 has any perfect cube divisors. We note that 24=8Γ—3. Thus, βˆ’10√24=βˆ’10ο€»βˆš8Γ—3=βˆ’10ο€»βˆš8Γ—βˆš3=βˆ’10ο€»2Γ—βˆš3=βˆ’20√3.

Now, 3 has no perfect cube divisors greater than 1, so we cannot simplify this term any further.

Next, we check βˆ’648 for any perfect cube divisors. Since this number is negative, we want a negative perfect cube. We note that βˆ’648=(βˆ’8)Γ—81. Hence, 5βˆšβˆ’648=5ο€»βˆšβˆ’8Γ—βˆš81=5ο€»βˆ’2√81=βˆ’10√81.

We want the radicand of the cube root to be the smallest positive integer possible. We can do this by checking to see if 81 has any perfect cube divisors. We note that 81=27Γ—3. Thus, βˆ’10√81=βˆ’10ο€»βˆš27Γ—βˆš3=βˆ’10ο€»3√3=βˆ’30√3.

Finally, we check 375 for any perfect cube divisors. We see that 375=125Γ—3. Thus, √375=√125Γ—βˆš3=5√3.

Substituting these values into the given expression yields βˆ’5√192+5βˆšβˆ’648+√375=βˆ’20√3βˆ’30√3+5√3.

We can then take out the shared factor of √3 to get βˆ’20√3βˆ’30√3+5√3=(βˆ’20βˆ’30+5)√3=βˆ’45√3.

In our final example, we will evaluate an algebraic expression by using the properties of the cube root.

Example 6: Evaluating an Algebraic Expression Involving Cube Roots

If π‘₯=√375 and 𝑦=√81, find the value of (π‘₯+𝑦).

Answer

We start by noting that we cannot directly add these radicals together, so instead, we look to simplify each cube root separately.

We need to identify a perfect cube that divides 375; we see that 375=125Γ—3. Hence, by using the fact that οŽ’οŽ’οŽ’βˆšπ‘ŽΓ—π‘=βˆšπ‘ŽΓ—βˆšπ‘ for any real numbers π‘Ž and 𝑏, we have √375=√125Γ—βˆš3=5√3.

Now, we look for a perfect cube that divides 81; we see that 81=27Γ—3. Hence, √81=√27Γ—βˆš3=3√3.

We can then substitute π‘₯=5√3 and 𝑦=3√3 into the expression to get (π‘₯+𝑦)=ο€»5√3+3√3.

We can take out the shared factor of √3 to get ο€»5√3+3√3=ο€»(5+3)√3=ο€»8√3.

Finally, we evaluate the exponent as follows: ο€»8√3=8√3Γ—8√3Γ—8√3=(8Γ—8Γ—8)Γ—ο€»βˆš3Γ—βˆš3Γ—βˆš3=512Γ—3=1536.

In the previous example, we may have been tempted to add √375 and √81 by adding their radicands to get √375+81. However, we cannot add radical expressions in this manner. An easy way to see this is to consider √8+√8. We know that √8=2, so √8+√8=2+2=4.

If we added the radicands, then we would get √16, which is not equal to 4.

In a similar way, we can show that (π‘₯+𝑦) is not the same as π‘₯+π‘¦οŠ©οŠ©. So, we cannot just evaluate the expression by adding the radicands: ο€»βˆš375+√81≠375+81.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • For any two real numbers π‘Ž and 𝑏, we have that οŽ’οŽ’οŽ’βˆšπ‘ŽΓ—βˆšπ‘=βˆšπ‘ŽΓ—π‘.
  • For any two real numbers π‘Ž and 𝑏, where 𝑏≠0, we have that οŽ’οŽ’οŽ’ο„žπ‘Žπ‘=βˆšπ‘Žβˆšπ‘.
  • If 𝑐 is an integer, we can use these results to write οŽ’οŽ’βˆšπ‘=π‘Žβˆšπ‘, where 𝑏 is smallest positive integer and has no perfect cube divisors greater than 1. This is called the simplest form of οŽ’βˆšπ‘.

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