Algebraic Identities: (x+y)³ & (x-y)³
The expressions shown above are called cubes of binomials.
Let's learn to expand these expressions:
$(x+y)_{3}(x−y)_{3}$
The cube of binomial can be written as shown.
Since,
$a_{3}=a×a_{2}$
The square in RHS is expanded further by using the identity:
$(x+y)_{2}=x_{2}+2xy+y_{2}$
The product on RHS is evaluated using the distributive property.
Then the like terms in RHS can be added.
Thus,
$(x+y)_{3}=x_{3}+3x_{2}y+3xy_{2}+y_{3}$
Let’s verify this identity.
In the LHS of the identity,
we put
$x=2&y=3$
Similarly in the RHS of the identity,
we put
$x=2&y=3$
Thus, the identity is verified.
Now, let's expand the second Identity
If we replace
$y$
with
$(−y)$
the expression changes to
$(x−y)_{3}$
So to find the expansion of
$(x−y)_{3}$
,
we can replace
$y$
with
$(−y)$
in
$(x+y)_{3}=x_{2}+3x_{2}y+3xy_{2}+y_{3}$
This is the required expansion for
$(x−y)_{3}$
Let’s now use these identities to factorize polynomials.
To factorize this polynomial, it can be compared with the expansion of either
$(x+y)_{3}$
or
$(x−y)_{3}$
Since all the terms of the polynomial are positive, it is compared with the expansion of
$(x+y)_{3}$
Terms of polynomials are rearranged to compare with the terms in the identity.
Then terms that are perfect cubes are identified.
Comparing the polynomial with the identity we have,
$x=3a&y=b$
Using the values of
$x&y$
, other terms of the polynomials are written as shown.
Since,
$x_{3}+3x_{2}y+3xy_{2}+y_{3}=(x+y)_{3}$
Let's factorize another polynomial.
This has both positive and negative terms, so it can be compared with the expansion of
$(x−y)_{3}$
The terms of polynomials are rearranged.
Then terms that are perfect cubes are identified.
Comparing the polynomial with the identity we have,
$x=2a&y=3b$
Using the values of
$x&y$
, other terms of the polynomials are written as shown.
Since
$x_{3}−3x_{2}y+3xy_{2}−y_{3}=(x−y)_{3}$
Revision
Expansion of cube of binomial
$(x+y)$
Replacing
$y$
with
$(−y)$
in the identity,
$(x+y)_{3}=x_{3}+3x_{2}y+3xy_{2}+y_{3}$
The End