Proof of Identity (a+b)(a-b) — lesson. Maths CBSE Live product, Class 8.

General expression for \((a+b)(a-b)\):

Using the distributive property of multiplication, we expand the expression \((a+b)(a-b)\) as follows:

\((a+b)(a-b)\) \(=\) \((a+b)a - (a+b)b\)

\(=\) \(a^2 + ba - ab -b^2\)

\(=\) \(a^2 + ab - ab -b^2\) [Since \(ba = ab\)]

\(=\) \(a^2 - b^2\)

Therefore, \((a+b)(a-b)\) \(=\) \(a^2 - b^2\).

Identity :

(a + b)(a - b)= a^{2} - b^{2}

Geometrical proof of identity:

Now, we construct a figure to understand the concept.

Then we construct a rectangle using the above information.

In the given figure, \(AB = AD = a\).

So, the area of square \(ABCD = a^2\).

Also, \(SB = DP = b\). Then the area of the rectangle \(SBCT = ab\).

Similarly, the area of the rectangle \(DPRC = ab\). And, the area of the square \(TQRC = b^2\).

Area of the rectangle \(DPQT = ab − b^2\).

Hence, \(\text{the area of the rectangle APQS = The area of square ABCD}\) \(\text{– area of rectangle STCB}\) \(\text{+ area of rectangle DPQT}\).

\begin{array}{l} = a^{2} - ab + (ab - b^{2}) \\ = a^{2} - ab + ab - b^{2} \\ = a^{2} - b^{2} \end{array}

Therefore,

(a + b)(a - b)= a^{2} - b^{2}

Example:

Simplify

(4 x + 8)(4 x - 8)

using the identity.

First, develop the given

(4 x + 8)(4 x - 8)

expression using the identity

(a + b)(a - b)= a^{2} - b^{2}

Here, \(a = 4x\); \(b = 8y\).

\begin{array}{l} (4 x + 8)(4 x - 8) = {(4 x)}^{2} - {(8)}^{2} \\ = 4^{2} \times x^{2} - {(8)}^{2} \\ = 16 x^{2} - 64 \end{array}

Therefore,

(4 x + 8)(4 x - 8)

\(=\) 16\(x^2 -\)64.

Did you find an error?

1. Proof of Identity (a+b)(a-b)