Algebraic Identities Of x³+y³ and x³-y³.
Suppose we have two cubes.
Let the volume of the first cube be
$x_{3}$
and the volume of second cube
$y_{3}$
.
Let’s join the cube side by side.
So, the total volume of the two cubes is,
We already have an identity for
$(x+y)_{3}$
.
So, let’s try to derive the identity
$x_{3}+y_{3}$
using the identity for
$(x+y)_{3}$
.
Let’s first try to understand this geometrically.
Let’s join our cubes as shown above.
We arranged both cubes in such a way to convert it into a cube as shown above.
Now, consider two cuboids of volumes as shown above.
Using these cuboids, let’s convert the volume of
$x_{3}+y_{3}$
into
$(x+y)_{3}$
. We will require three of both the cuboids.
We can get the cube as,
We can interpret this mathematically as,
The required volume thus can be given as,
So, let’s try to interpret and verify this mathematically.
We know the algebraic expansion of
$(x+y)_{3}$
.
Rearranging the terms in the expansion, we will get our identity for
$x_{3}+y_{3}$
.
Thus, we have verified our identity mathematically.
Again, if we replace
$x$
with
$−y$
in the expression, we have
This identity can also be expressed as,
This is the required standard algebraic identity.
In the expression, if we replace
$y$
with
$(−y)$
, we will get the identity
$x_{3}−y_{3}$
.
Now, let’s further verify this numerically with an example.
To verify, let’s take the values for
$x$
and
$y$
and put in the LHS and RHS of the identity.
On putting the value, the LHS will give the value,
And on putting the values in the RHS, we will obtain,
As, we get the same result for LHS and RHS, our identity is verified.
Now, let’s apply this concept in solving the factorisation problems in polynomials.
Consider we have a polynomial as given above.
We have determined our standard identity as,
To use our standard identity, let’s assume,
Substituting the values of
$x$
and
$y$
in the formula,
Thus, we have factorized a given polynomial using our algebraic identity.
Revision
One of the important algebraic identity can be given by,
With the use of this identity, we can factorise any polynomial.
The End