Algebraic identities are relationships that are always true. Let us begin with an example.
We can start by considering the expression $a^2 + ab$
Taking $a$ common from both terms we get the expression $a(a + b)$
Therefore, when we write:
$a^2 + ab = a(a + b)$
this is true for all values of $a$ and $b$. These relationships that are true for all values of its variables, are called $\underline{Identities}$
Identities are different from equations, in the sense that while identities are always true, equations are true only for one or more specific values of the unknown.
For example the equation $3x + 2 = 11$ is valid only for $x = 3$
Similarly, the equation $2x^2 - 50 = 0$ is valid only for $x = 5$ or $x = -5$
Therefore, these are equations and not identities.
Now let us look at some common form of identities.
II. Identities Of The Form $(a \pm b)^2$
In this section we will look at the derivation of two identities, namely:
$(a + b)^2 = a^2 + 2ab + b^2$
and
$(a - b)^2 = a^2 - 2ab + b^2$
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We will derive these identities by multiplying and expanding the $LHS$ of the identities.
$(a + b)^2$
$= (a + b)(a + b)$
$= a \times a + a \times b + b \times a + b \times b$
$= a^2 + ab + ab + b^2$
$= a^2 + 2ab + b^2$
$\therefore (a + b)^2 = a^2 + 2ab + b^2$
Similarly,
$(a - b)^2$
$= a \times a - a \times b - b \times a + b \times b$
$= a^2 - ab - ab + b^2$
$= a^2 - 2ab + b^2$
$\therefore (a - b)^2 = a^2 - 2ab + b^2$
III. Identities Of The Form $(a^2 - b^2)$
Now, let us consider the identity $a^2 - b^2$
We can factorise this expression using the following steps:
$a^2 - b^2$
= $a^2 - ab + ab - b^2$ (Observe that we have added and subtracted the same term $ab$, so our expression remains the same)
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= $a(a - b) + b(a-b)$
= $(a+b)(a-b)$
So, we can see that an algebraic expression of the form $a^2 - b^2$ is equal to $(a+b)(a-b)$.
Let us take a few examples to understand this.
We will factorise the polynomial $x^2 - 9$
We can write $x^2 - 9 = x^2 - 3^2 = (x+3)(x-3)$
This works even if one of the terms is not a perfect square, like the polynomial $9x^4 - 32$
We can see that $32$ is not a perfect square but we can still write it as $(4\sqrt{2})^2$
So, we can write,
$9x^4 - 32$
$=(3x^2)^2 - (4\sqrt{2})^2$
$=(3x^2 - 4\sqrt{2})(3x^2 + 4\sqrt{2})$
And, that gives us the factors for our given polynomial.
Now we will see another common form of expression which is $a^3 - b^3$. We can factorise this expression using the following steps:
$a^3 - b^3$
$= a^3 - a^2b + a^2b - ab^2 + ab^2 - b^3$
$= a^2(a-b) + ab(a-b) + b^2(a-b)$
$= (a-b)(a^2 + ab + b^2)$
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Let us take an example by factorising the expression $8x^3 - 27$
$8x^3 - 27$
$= (2x)^3 - 3^3$
We can apply our formula for $a^3 - b^3$ using $a = 2x$ and $b = 3$. Therefore, we get:
$(2x)^3 - 3^3$
$= (2x-3)\left[(2x)^2 + 2x\times3 + (3)^2\right]$
$=(2x-3)(4x^2 + 6x + 9)$
Now we will take a look at another common form of algebraic expression, that can be factorised very easily.
If our expression is of the form, $a^3 + b^3$, we can factorise them very easily. Let's see how.
$a^3 + b^3$
$=a^3+a^2b-a^2b+ab^2-ab^2+b^3$
$=a^2(a+b)-ab(a+b)+b^2(a+b)$
$=(a+b)(a^2-ab+b^2)$
Now let us take an example to illustrate the application of this. We will factorise $27p^3 + 125$
$27p^3 + 125$
$= (3p)^3 + 5^3$
$= (3p + 5)\{(3p)^2 + (3p)\times(5) + (5)^2\}$
$= (3p + 5)(9p^2 + 15p + 25)$
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IV. Quick Recap
A quick recap of the three formulae that we learnt today: