Math Doubts

$x^3-y^3$ Identity

Formula

$x^3-y^3$ $\,=\,$ $(x-y)(x^2+y^2+xy)$

Introduction

Let’s assume, $x$ and $y$ denote two quantities in algebraic form.

  1. The subtraction of $y$ from the $x$ is written as $x-y$.
  2. The subtraction of $y$ cubed from $x$ cubed is written as $x^3-y^3$.
  3. Add the sum of the squares of $x$ and $y$ to the product of $x$ and $y$, and their addition is written as $x^2+y^2+xy$.

According to the difference of cubes arithmetic property, the subtraction of $y$ cube from $x$ cube is equal to the product of the subtraction of $y$ from $x$ and the addition of the sum of squares of $x$ and $y$ and the product of $x$ and $y$.

$\therefore\,\,\,$ $x^3-y^3$ $\,=\,$ $(x-y)$ $\times$ $(x^2+y^2+xy)$

Now, let us learn much more about the $x$ cube minus $y$ cube algebraic identity.

Other form

The $x$ cube minus $y$ cube algebraic identity is alternatively written in the following mathematical form.

$a^3-b^3$ $\,=\,$ $(a-b)(a^2+b^2+ab)$

Uses

The $x$ cubed minus $y$ cubed algebraic identity is mainly used in two different cases.

  1. It is used as a formula to find the difference of cubes of two quantities in basic mathematics.
  2. It is also used as a formula to express the difference of two quantities in mathematical form as a product of two factors.

Proofs

The $x$ cube minus $y$ cube algebraic identity can be derived mathematically in two methods.

Algebraic Method

Learn how to prove the $x$ cubed minus $y$ cubed identity by the algebraic identities.

Geometric Method

Learn how to derive the $x$ cube minus $y$ cube formula geometrically by the volume of a cube.

Verification

Assume that $x = 3$ and $y = 1$

$(1).\,\,$ $x-y$ $\,=\,$ $3-1$ $\,=\,$ $2$

$(2).\,\,$ $x^2+y^2+xy$ $\,=\,$ $3^2+1^2+3 \times 1$ $\,=\,$ $9+1+3$ $\,=\,$ $13$

$(3).\,\,$ $x^3-y^3$ $\,=\,$ $3^3-1^3$ $\,=\,$ $27-1$ $\,=\,$ $26$

Now, find the product of the $x-y$ and $x^2+y^2+xy$.

$\implies$ $(x-y)$ $\times$ $(x^2+y^2+xy)$ $\,=\,$ $2 \times 13$ $\,=\,$ $26$

It is evaluated that the product of them is equal to $26$, and the difference of cubes of $x$ and $y$ is also equal to $26$.

$\,\,\,\therefore\,\,\,\,\,\,$ $x^3-y^3$ $\,=\,$ $26$ $\,=\,$ $(x-y)(x^2+y^2+xy)$

In this way, you can verify the $x$ cube minus $y$ cube algebraic identity by taking any two numbers.

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