Cubes & Cube Roots Class 8: Patterns, Factorization & Estimation

You already know that $5^2 = 25$ gives you the area of a square with side $5$. Now imagine a cube with side $5$. Its volume? That's $5^3 = 5 \times 5 \times 5 = 125$ cubic units. That's what cubing a number means, multiplying it by itself three times.

Cubes and cube roots are covered in NCERT Class 8 Math (Chapter 6: Cubes and Cube Roots) and they build directly on what you learned about squares. If you're comfortable with squares, cubes will feel like a natural extension. Let's get into it!

A number $n$ is a perfect cube if $n = m^3$ for some integer $m$. Here are the cubes of the first ten natural numbers:

Memorising these first ten cubes is extremely helpful for quick calculations in exams. You'll recognise perfect cubes instantly and save valuable time!

Just like squares, cubes have some beautiful patterns that make them easier to understand and remember.

To determine whether a number is a perfect cube, use prime factorization. If every prime factor appears a number of times that is a multiple of $3$, then the number is a perfect cube.

Example: Is $5832$ a perfect cube?

Both $2$ and $3$ appear a multiple-of-three times ($3$ and $6$ respectively). So yes, $5832$ is a perfect cube.

Example: Is $392$ a perfect cube?

Here $7$ appears only twice (not a multiple of $3$). So $392$ is not a perfect cube. To make it one, multiply by $7$: $392 \times 7 = 2744 = 14^3$.

This is the main method for finding cube roots in CBSE Class 8. The steps are:

1. Find the complete prime factorization.
2. Group the prime factors into triplets (groups of three).
3. Take one factor from each triplet and multiply them.

Example 1: Find $\sqrt[3]{13824}$.

Example 2: Find $\sqrt[3]{46656}$.

Example 3: Find the smallest number by which $1296$ must be multiplied to obtain a perfect cube.

For a perfect cube, we need each exponent to be a multiple of $3$. We need $2^2$ more of $2$ (to make $2^6$) and $3^2$ more of $3$ (to make $3^6$). So multiply by $2^2 \times 3^2 = 4 \times 9 = 36$.

This is where cube roots differ fundamentally from square roots. While square roots of negative numbers don't exist in real numbers, cube roots of negative numbers are perfectly fine!

Why? Because a negative number multiplied by itself three times gives a negative result:

So $\sqrt[3]{-27} = -3$.

The general rule is:

Examples:

This makes sense intuitively: cubing preserves the sign (positive stays positive, negative stays negative), so cube rooting does too.

For larger numbers where prime factorization might be tedious, you can estimate cube roots using the memorised cubes.

Example: Estimate $\sqrt[3]{79507}$.

Step 1: Look at the last three digits: $507$. Since $7^3 = 343$ and a number ending in $3$ gives a cube ending in $7$, the units digit of the cube root is $3$.

Step 2: Ignore the last three digits and look at $79$. Now $4^3 = 64$ and $5^3 = 125$. Since $64 < 79 < 125$, the tens digit is $4$ (we take the smaller one).

So $\sqrt[3]{79507} \approx 43$. Let's verify: $43^3 = 79507$. Spot on!

This estimation method works perfectly for perfect cubes and gives good approximations for non-perfect cubes too.

Here's your quick-reference summary:

• A perfect cube $n = m^3$ has every prime factor appearing a multiple of $3$ times.
  - Cube roots of negative numbers exist: $\sqrt[3]{-a} = -\sqrt[3]{a}$.
  - Use prime factorization to find cube roots: group factors into triplets, pick one from each.
  - The sum of cubes formula: $1^3 + 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$.
  - Cubes can be expressed as sums of consecutive odd numbers.
  - The estimation method uses last-digit patterns and memorised cubes to quickly approximate cube roots.

Ready to put this knowledge to the test? Jump into SparkEd's interactive practice and work through cubes and cube roots problems with step-by-step guidance!