One of the most frequent questions in algebra involves finding the difference of cubes when given a simple linear relationship.
The Problem:
If $x – \frac{1}{x} = 5$, find the value of $x^3 – \frac{1}{x^3}$.
The Secret Identity:
To solve this, we use the identity for the cube of a binomial:
The Step-by-Step Solution:
Start with the given: $x – \frac{1}{x} = 5$
Cube both sides: $(x – \frac{1}{x})^3 = 5^3$
Expand the left side: $x^3 – \frac{1}{x^3} – 3(x)(\frac{1}{x})(x – \frac{1}{x}) = 125$
Final Calculation: $x^3 – \frac{1}{x^3} – 15 = 125 \rightarrow \mathbf{140}$
When you see a binomial multiplied by a trinomial, don’t rush into long multiplication! It might be a hidden “Difference of Cubes.”
The Problem:
Find the product: $(4x – 5y)(16x^2 + 20xy + 25y^2)$
The Identity:
The Solution:
If we let $a = 4x$ and $b = 5y$:
$a^2 = (4x)^2 = 16x^2$
$ab = (4x)(5y) = 20xy$
$b^2 = (5y)^2 = 25y^2$
Since the expression fits the pattern perfectly, we can jump straight to the answer:
Finally, let’s look at one of the more advanced identities used in higher-level algebra.
The Identity:
Why it works:
If you take the Right Hand Side (R.H.S) and apply the distributive law (multiplying every term in the first bracket by every term in the second), most terms will cancel each other out, leaving you exactly with the Left Hand Side (L.H.S).
Pro Tip: This identity is incredibly useful in geometry and physics problems where you need to simplify complex polynomials quickly.
| Name | Formula |
| Difference of Cubes | $a^3 – b^3 = (a – b)(a^2 + ab + b^2)$ |
| Cube of Difference | $(a – b)^3 = a^3 – b^3 – 3ab(a – b)$ |
| Special Cubic Identity | $a^3 + b^3 + c^3 – 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ |
The Problem:
Find the product $(x+y+2z)(x^2+y^2+4z^2-xy-2yz-2zx)$.
The Solution:
This expression matches the structure of the identity $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.
Here, $a=x$, $b=y$, and $c=2z$.
Final Answer: $x^3+y^3+8z^3-6xyz$.
You don’t always need to know the value of $a$, $b$, and $c$ individually to find the value of their cubic sum.
The Problem:
If $a+b+c=6$ and $ab+bc+ca=11$, find the value of $a^3+b^3+c^3-3abc$.
The Key Transformation:
The standard identity can be rewritten to use the information we actually have:
The Calculation:
Substitute the known values: $6[(6)^2 – 3(11)]$.
Simplify inside the brackets: $6[36 – 33]$.
Final result: $6(3) = \mathbf{18}$.
Keep these formulas handy for your homework and exams:
| Type | Formula |
| Sum Cubed | $(a+b)^3 = a^3+b^3+3ab(a+b)$ |
| Difference Cubed | $(a-b)^3 = a^3-b^3-3ab(a-b)$ |
| Sum of Cubes | $a^3+b^3 = (a+b)(a^2-ab+b^2)$ |
| Difference of Cubes | $a^3-b^3 = (a-b)(a^2+ab+b^2)$ |
| The Special Case | If $a+b+c=0$, then $a^3+b^3+c^3 = 3abc$ |
Expand: $(2x + \frac{3}{x})^3$.
Using the $(a+b)^3$ identity:
$(2x)^3 + (\frac{3}{x})^3 + 3(2x)(\frac{3}{x})(2x + \frac{3}{x})$
The $x$ terms in the middle cancel out: $3(2)(3) = 18$.
Result: $8x^3 + \frac{27}{x^3} + 18(2x + \frac{3}{x})$.
Building on our previous exploration of algebraic identities, let’s look at more complex applications and a handy “cheat sheet” of cubic formulas. Mastering these allows you to solve multi-variable equations with speed and precision.
Sometimes you are faced with a product involving three variables ($x$, $y$, and $z$). Instead of multiplying every term individually, you can recognize the structure of a specific identity.
The Problem:
Find the product $(x+y+2z)(x^2+y^2+4z^2-xy-2yz-2zx)$.
The Solution:
This expression matches the structure of the identity $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.
In this case, $a=x$, $b=y$, and $c=2z$.
We can rewrite the expression as $(x)^3+(y)^3+(2z)^3-3(x)(y)(2z)$.
Final Answer: $x^3+y^3+8z^3-6xyz$.
You don’t always need to know the individual values of variables to find the value of their cubic sum.
The Problem:
If $a+b+c=6$ and $ab+bc+ca=11$, find the value of $a^3+b^3+c^3-3abc$.
The Key Transformation:
The standard identity can be modified to use the information provided:
$a^3+b^3+c^3-3abc = (a+b+c)[(a+b+c)^2 – 3(ab+bc+ca)]$.
The Calculation:
Substitute the known values: $6[(6)^2 – 3(11)]$.
Simplify inside the brackets: $6[36 – 33]$.
Final result: $6(3) = \mathbf{18}$.
Keep these formulas handy for your homework and exams:
| Name/Type | Formula |
| Sum Cubed | $(a+b)^3 = a^3+b^3+3ab(a+b)$ |
| Difference Cubed | $(a-b)^3 = a^3-b^3-3ab(a-b)$ |
| Sum of Cubes | $a^3+b^3 = (a+b)(a^2-ab+b^2)$ |
| Difference of Cubes | $a^3-b^3 = (a-b)(a^2+ab+b^2)$ |
| Special Cubic Identity | $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ |
| The Zero Condition | If $a+b+c=0$, then $a^3+b^3+c^3 = 3abc$ |
Identities aren’t just for variables; they make mental math much easier.
Large Number Cubing: To evaluate $(1002)^3$, use $(1000+2)^3$. By applying the $(a+b)^3$ identity, the result is $1,006,012,008$.
Fractional Expansion: When expanding $(2x + \frac{3}{x})^3$, use the $(a+b)^3$ formula. The $x$ and $\frac{1}{x}$ terms cancel in the middle, resulting in $8x^3 + \frac{27}{x^3} + 36x + \frac{54}{x}$.
Evaluating with Zero Sums: To find $30^3 + 20^3 – 50^3$, notice that $30+20+(-50)=0$. Using the “Zero Condition” identity, the answer is simply $3(30)(20)(-50) = \mathbf{-90,000}$.
The Identity:
When you see a product like $(x+y+2z)(x^2+y^2+4z^2-xy-2yz-2zx)$, don’t use long multiplication. Match it to the identity where $a=x, b=y,$ and $c=2z$.
Result: $x^3 + y^3 + 8z^3 – 6xyz$.
You can find the value of $a^3+b^3+c^3-3abc$ even if you only know the sums of the variables.
Given: $a+b+c=6$ and $ab+bc+ca=11$.
Formula Shift: Use $(a+b+c)[(a+b+c)^2 – 3(ab+bc+ca)]$.
Calculation: $6[(6)^2 – 3(11)] = 6[36-33] = 18$.
A unique property of algebra is that if $a+b+c=0$, then $a^3+b^3+c^3 = 3abc$. This makes evaluating large numbers incredibly fast.
Example: Evaluate $30^3 + 20^3 – 50^3$.
Observation: Since $30 + 20 + (-50) = 0$, we just calculate $3(30)(20)(-50)$.
Answer: $-90,000$.
When working with expressions like $(2x + \frac{3}{x})^3$, we use the standard $(a+b)^3$ expansion.
Cube the terms: $(2x)^3 + (\frac{3}{x})^3$.
Add the middle term: $3(2x)(\frac{3}{x})(2x + \frac{3}{x})$.
Simplify: The $x$ and $\frac{1}{x}$ cancel out, leaving $18(2x + \frac{3}{x})$.
Final Form: $8x^3 + \frac{27}{x^3} + 36x + \frac{54}{x}$.
| Identity Type | Formula |
| Sum Cubed | $(a+b)^3 = a^3+b^3+3ab(a+b)$ |
| Difference Cubed | $(a-b)^3 = a^3-b^3-3ab(a-b)$ |
| Sum of Cubes | $a^3+b^3 = (a+b)(a^2-ab+b^2)$ |
| Difference of Cubes | $a^3-b^3 = (a-b)(a^2+ab+b^2)$ |
The Core Identity:
You can find the value of this complex expression even when individual values for $a, b,$ and $c$ are unknown, provided you have their sums and products.
Problem: If $x + y + z = 8$ and $xy + yz + zx = 20$, find $x^3 + y^3 + z^3 – 3xyz$.
The Shortcut Formula: $a^3 + b^3 + c^3 – 3abc = (a + b + c)[(a + b + c)^2 – 3(ab + bc + ca)]$.
The Calculation: $8[8^2 – 3(20)] = 8[64 – 60] = 8(4) = \mathbf{32}$.
A unique property of these identities is that if the sum of three variables is zero ($a + b + c = 0$), then the sum of their cubes is equal to three times their product.
The Identity: If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$.
Mental Math: To evaluate $30^3 + 20^3 – 50^3$, notice that $30 + 20 + (-50) = 0$. The answer is simply $3(30)(20)(-50) = \mathbf{-90,000}$.
Simplifying Fractions: The expression $\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}$ can be rewritten as $\frac{a^3 + b^3 + c^3}{abc}$. If $a+b+c=0$, this simplifies to $\frac{3abc}{abc}$, which equals 3.
When dealing with reciprocal terms or larger coefficients, standard cubic expansions still apply.
Reciprocal Power Shifts: Starting from $x^4 + \frac{1}{x^4} = 47$, you can derive $x + \frac{1}{x} = \pm 3$. Using the identity $(a+b)^3$, you can then find that $x^3 + \frac{1}{x^3}$ is either 18 or -18.
Complex Binomial Cubes: Expanding $(2x + \frac{3}{x})^3$ involves cubing the individual terms and adding the middle product: $(2x)^3 + (\frac{3}{x})^3 + 3(2x)(\frac{3}{x})(2x + \frac{3}{x})$, which simplifies to $8x^3 + \frac{27}{x^3} + 36x + \frac{54}{x}$.
| Name | Formula |
| Sum Cubed | $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$ |
| Difference Cubed | $(a-b)^3 = a^3 – b^3 – 3ab(a-b)$ |
| Sum of Cubes | $a^3 + b^3 = (a+b)(a^2 – ab + b^2)$ |
| Difference of Cubes | $a^3 – b^3 = (a-b)(a^2 + ab + b^2)$ |