Difference of Squares, Sum and Difference of Cubes

Finding the Difference of Squares, Sum and Difference of Cubes of two numbers involves lengthy calculations. At first, we must find the squares of two numbers, then only we can find the difference of the results of some of the results. If we need to find the sum or difference of two squares and two cubes using two large numbers, we should use algebraic identities to avoid lengthy and complicated calculations.

In factorisation of algebraic expressions also, we use algebraic identities such as sum or difference of squares, sum and difference of cubes wherever required. Let us learn about the difference between squares, sum, and cubes using

algebraic identities and their applications.

Algebraic Identities

Algebraic identities are algebraic equations, which is valid for all the values of variables in them. Algebraic equations are mathematical expressions consisting of numbers, variables (unknown values) and mathematical operators (addition, subtraction, multiplication, division, etc.)

We use algebraic identities for the factorisation of polynomials. In this way, algebraic identities are used in the evaluation of algebraic expressions and solving different polynomials. These are primarily used to determine the factors of polynomials. Algebraic identities are used in different branches of mathematics, such as algebra, geometry, calculus, trigonometry etc.

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Factoring Algebraic Identities

Factoring algebraic expressions are made much easier by algebraic identities. To derive some of the higher algebraic identities such as \({a^4} – {b^4},\) the basic algebraic identities \({a^2} – {b^2}\) must be known. The following is a list of algebraic identities that can be used to factor polynomial expressions.

1. Identity -1: \({a^2} – {b^2} = \left({a + b} \right)\left({a – b} \right)\)

2. Identity-2: \({a^3} – {b^3} = \left({a – b} \right)\left({{a^2} + ab + {b^2}}\right)\)

3. Identity-3: \({a^3} + {b^3} = \left({a + b} \right)\left({{a^2} – ab + {b^2}} \right)\)

4. Identity-4: \({a^4} – {b^4} = \left({{a^2} -{b^2}}\right)\left({{a^2} + {b^2}} \right)\)

Difference of Squares

The difference of squares formula is an algebraic identity used to find the difference of squares between two numbers without calculating the squares themselves. The difference of squares formula is used to factorise quadratic binomials. The formula for calculating the difference between two squares is:

Let us understand through an example.

Example 1: Let us simplify \({x^2} – 25.\)
Step 1: In the above algebraic expression, the square root of \({x^2}\) is \(x,\) and the square root of \(25\) is \(5\) as \({5^2} = 25\)
So, we can write \({x^2} – 25\) as \({\left( x \right)^2} – {\left( 5 \right)^2}.\)
Step 2: Let us assume \(x\) as \(a\) and \(5\) as \(b.\)
Step 3: Now, according to the formula, we can write \(\left({x + 5} \right)\left({x – 5} \right)\) by substituting \(a = x\) and \(b = 5\) in \(\left({a + b} \right)\left({a – b} \right).\)
Hence, \({x^2} – 25 = \left({x + 5} \right)\left({x – 5} \right)\)

Example 2: Evaluate \(101 \times 99\)
We can evaluate the product of these numbers using algebraic identities.
Step 1: We can write \(101 \times 99\) as \(\left({100 + 1} \right)\left({100 – 1} \right).\)
Step 2: Let us assume \(a = 100\) and \(b = 1\)
Step 3: Now, according to the formula, we can write \(\left({100 + 1} \right)\left({100 – 1} \right) = {100^2} – {1^2}\) by substituting \(a = 100\) and \(b = 1\) in \(\left({a + b} \right)\left({a – b} \right) = {a^2} – {b^2}.\)
Hence, \(101 \times 99 = 10000 – 1 = 9999\)

Sum of Cubes

The formula to determine the addition of two perfect cubes such as \({a^3} + {b^3}\) is known as the sum of cubes formula. This formula for factoring algebraic expressions of various types is beneficial. This formula is also easy to memorise and can be done in just a few minutes. The formula for the sum of cubes is:

Let us take some examples and understand.

Example 1: \({x^3} + 8\)
Step 1: In the above algebraic expression, the cube root of \({x^3}\) is \(x,\) and cube root of \(8\) is \(2\) as \({2^3} = 8\)
So, we can write \({x^3} + 8\) as \({x^3} + {2^3}.\)
Step 2: Let us assume \(x\) as \(a\) and \(2\) as \(b.\)
Step 3: Now, according to the formula, we can write \(\left({x + 2} \right)\left({{x^2} – 2x + {2^2}} \right)\) by substituting \(a = x\) and \(b = 2\) in \(\left({a + b} \right)\left({{a^2} – ab + {b^2}} \right).\)
Hence, \({x^3} + 8 = \left({x + 2} \right)\left({{x^2} – 2x + 4} \right)\)

Example 2: Evaluate \({20^3} + {15^3}\)
We can find the value for \({20^3} + {15^3}\) by finding the value of \({20^3}\) and \({15^3}\) separately then adding the results. It may take time as the calculation is lengthy. To reduce the complexity, we will apply here the formula of the sum of cubes.
Let us take \(a = 20\) and \(b = 15\)
Using the sum of cubes formula \({a^3} + {b^3} = \left({a + b}\right)\left({{a^2} – ab + {b^2}} \right)\)
We will substitute \(a = 20,b = 15\) in the \({a^3} + {b^3}\) formula
\({a^3} + {b^3} = \left({a + b}\right)\left({{a^2} – ab + {b^2}} \right).\)
\({20^3} + {15^3} = \left({20 + 15} \right)\left({{{20}^2} – 20 \times 15 + {{15}^2}} \right).\)
\( = \left({35} \right)\left({400 – 300 + 225} \right)\)
\( = \left({35} \right)\left({325} \right) = 11,375\)
Hence, \({20^3} + {15^3} = 11375\)

Difference of Cubes

The difference of cubes formula is used to find the difference of the cubes between two numbers without calculating the cubes themselves. It is an algebraic identity. The difference of cubes formula is used to factorise cubic binomials. The \({a^3} – {b^3}\) formula is another name for the difference of cubes formula. The formula for calculating the difference between two cubes is:

Example 1: \(27{x^3} – 125\)
Step 1: In the above algebraic expression, cube root of \(27{x^3}\) is \(3x\) as \({\left({3x} \right)^3} = 27{x^3}\) and cube root of \(125\) is \(5\) as \({5^3} = 125\)
So, we can write \(27{x^3} – 125\) as \({\left({3x} \right)^3} -{\left( 5 \right)^3}.\)
Step 2: Let us assume \(3x\) as \(a\) and \(5\) as \(b.\)
Step 3: Now, according to the formula, we can write \(\left({3x – 5} \right)\left\{{{{\left({3x} \right)}^2} + 3x \times 5 + {{\left( 5 \right)}^2}}\right\}\) by substituting \(a = 3x\) and \(b = 5\) in \(\left({a – b} \right)\left({{a^2} + ab + {b^2}} \right).\)
Hence, \(27{x^3} – 125 = \left( {3x – 5} \right)\left( {9{x^2} + 15x + 25} \right)\)

Example 2: Evaluate \({18^3} – {14^3}\)
We can find the value for \({18^3} – {14^3}\) by finding the value of \({18^3}\) and \({14^3}\) separately then subtracting the results. It may take time as the calculation is lengthy. To reduce the complexity, we will apply here the formula of the difference of cubes.
Let us assume \(a = 18\) and \(b = 14\)
Using the difference of cubes formula \({a^3} – {b^3} = \left({a – b} \right)\left({{a^2} + ab + {b^2}} \right).\)
We will substitute \(a = 18,b = 14\) in the \({a^3} -{b^3}\) formula
\({a^3} – {b^3} = \left({a – b} \right)\left({{a^2} + ab + {b^2}} \right).\)
\({18^3} – {14^3} = \left({18 – 14} \right)\left({{{18}^2} + 14 \times 18 + {{14}^2}} \right).\)
\( = \left( 4 \right)\left({324 + 252 + 196} \right)\)
\( = \left( 4 \right)\left({772} \right) = 3088\)
Hence, \({18^3} – {14^3} = 3088\)

Difference of Squares Sum and Difference of Cubes: Solved Examples

Q.1. Find the value of \(197 \times 203\) by using the standard algebraic identities.
Ans: Given \(197 \times 203\)
It can be written as \(\left({200 – 3} \right) \times \left({200 + 3} \right)\)
By using the algebraic identity for the difference of squares, we have \(\left({a + b} \right)\left({a – b} \right) = {a^2} – {b^2}\)
Substituting the value of \(a = 200\) and \(b = 3,\) we get
\(\left({200 – 3} \right) \times \left({200 + 3}\right) = {\left({200} \right)^2} -{3^2}\)
\(= 40000 – 9\)
\( = 39991\)
Hence, the value of \(197 \times 203 = 39991.\)

Q.2. Factorise \(\left({{x^4} – 1}\right)\)
Ans: Given: \(\left({{x^4} – 1}\right)\)
It can be written as \(\left[{{{\left({{x^2}} \right)}^2} – 1} \right]\)
By using the standard identity: \(\left({a + b} \right)\left({a – b} \right) = {a^2} – {b^2},\) we get, \(\left({{x^4} – 1} \right) = \left({{x^2} – 1} \right)\left({{x^2} + 1}\right)\)
\( = \left({x – 1} \right)\left({x + 1} \right)\left({{x^2} + 1} \right)\) (Again, by using the identity: \(\left({a + b} \right)\left({a – b} \right) = {a^2} – {b^2}\))
Hence, the factors of \(\left({{x^4} – 1} \right)\) are \(\left({x – 1} \right)\left({x + 1} \right)\left({{x^2} + 1}\right).\)

Q.3. Find the factors of the expression \(8{x^3} – 64{y^3}\)
Ans: Step 1: In the above algebraic expression, cube root of \(8{x^3}\) is \(2x\) as \({\left({2x} \right)^3} = 8{x^3}\) and cube root of \(64{y^3}\) is \(\left( 4 \right){y^3}\) as \(\left( 4 \right){y^3} = 64{y^3}\)
So, we can write \(8{x^3} – 64{y^3}\) as \({\left({2x} \right)^3} – {\left({4y} \right)^3}.\)
Step 2: Let us assume \(2x\) as \(a\) and \(4y\) as \(b.\)
Step 3: Now, according to the formula, we can write \(\left({2x – 4y} \right)\left\{{{{\left({2x} \right)}^2} + 2x \times 4y + {{\left({4y} \right)}^2}} \right\}\) by substituting \(a = 2x\) and \(b = 4y\) in \(\left({2x – 4y} \right)\left({4{x^2} + 8xy + 16{y^2}}\right).\)
Hence, \(8{x^3} – 64{y^3} = 2\left({x – 2y} \right)\left({2{x^2} + 4xy + 8{y^2}}\right)\)

Q.4. Evaluate \({32^3} +{17^3}\)
Ans: We can find the value for \({32^3} +{17^3}\) finding the value of \({32^3}\) and \({17^3}\) separately then adding the results. It may take time as the calculation is lengthy. To reduce the complexity, we will apply here the formula of the sum of cubes.
Let us assume \(a = 32\) and \(b = 17\)
Using the sum of cubes formula \({a^3} + {b^3} = \left({a + b} \right)\left({{a^2} – ab + {b^2}} \right).\)
We will substitute \(a = 32,b = 17\) in the \({a^3} + {b^3}\) formula
\({a^3} + {b^3} = \left({a + b} \right)\left({{a^2} – ab + {b^2}} \right).\)
\({32^3} + {17^3} = \left({32 + 17} \right)\left({{{32}^2} – 32 \times 17 + {{17}^2}}\right).\)
\( = \left({49} \right)\left({1024 – 544 + 289} \right)\)
\( = \left({49}\right)\left({769} \right) = 37,681\)
Hence, \({32^3} + {17^3} = 37,681\)

Q.5. Evaluate \({28^3} – {13^3}\)
Ans: We can find the value for \({28^3} – {13^3}\) by finding the value of \({28^3}\) and \({13^3}\) separately then subtracting the results. It may take time as the calculation is lengthy. To reduce the complexity, we will apply here the formula of the difference of cubes.
Let us assume \(a = 28\) and \(b = 13\)
Using the difference of cubes formula \({a^3} – {b^3} = \left({a – b} \right)\left({{a^2} + ab +{b^2}} \right).\)
We will substitute \(a = 28,b = 13\) in the \({a^3} – {b^3}\) formula
\({a^3} – {b^3} = \left({a – b} \right)\left({{a^2} + ab +{b^2}} \right).\)
\({28^3} – {13^3} = \left({28 – 13} \right)\left({{{28}^2} + 28 \times 13 + {{13}^2}} \right)\)
\( = \left({15} \right)\left({784 + 364 + 169} \right)\)
\( = \left({15} \right)\left({1317} \right) = 19755\)
Hence, \({28^3} – {13^3} = 19755\)

Summary

In this article, we learnt the algebraic identities for the difference between squares and sum and the difference between cubes. We discussed the formulas of the difference of cubes and sum of cubes, the difference of squares. We also solved some examples using these formulas.

Learn the Concepts on Cube and Cuboid

FAQs

Q.1. What are the formulas of the difference of the squares and cubes?
Ans: Algebraic identities are the formulas used to find the difference of squares and cubes. The formula of the difference of the squares is,
\({a^2} – {b^2} = \left({a + b} \right)\left({a – b} \right)\)
The formula of the difference of the cubes is,
\({a^3} – {b^3} = \left({a – b} \right)\left({{a^2} + ab + {b^2}} \right)\)

Q.2. How do you find the sum and difference of cubes?
Ans: The formula to determine the addition of two perfect cubes such as \({a^3} + {b^3}\) is known as the sum of cubes formula.
Formula: \({a^3} + {b^3} = \left({a + b}\right)\left({{a^2} – ab + {b^2}} \right)\)
The difference of cubes formula is used to find the difference of the cubes between two numbers without calculating the cubes themselves.
Formula:
\({a^3} -{b^3} = \left({a – b} \right)\left({{a^2} + ab + {b^2}}\right)\)

Q.3. What is the example of the sum and difference of two cubes?
Ans: An example of the sum of two cubes is \({8^3} + {4^3}.\) Here, we need not calculate the value of \({8^3}\& {4^3}\) to get the value of the sum. Instead, we can use the sum of cubes formula to solve them.
An example of the difference of the two cubes is \({8^3} – {4^3}.\) We can use the difference of cubes formula here to simplify this expression instead of solving it after finding the value of \({8^3}\,\& \,{4^3}.\)

Q.4. What are squares and cubes in math?
Ans: A number multiplied by itself is called the square of the number; as a result, \({4^2}\) is equal to \(16,\) and \(2\) squared is equal to \({2^2},\) which equals four.
The cube of a number is that number multiplied by itself three times. In other words, it means a number multiplied by itself, and the result will be again multiplied by the number itself. Example, \({4^3}\) is \(4 \times 4 \times 4 = 64.\)

Q.5. What is the formula of the sum of two squares?
Ans: The formula of the sum of two squares is \({a^2} + {b^2} = {\left({a + b}\right)^2} – 2ab\)

We hope this detailed article on the difference of squares, sum and difference of cubes helped you in your studies. If you have any doubts, queries or suggestions regarding this article, feel to ask us in the comment section and we will be more than happy to assist you. Happy learning!