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When it comes to solving the algebraic equation, identities play a vital role to solve them. In this lesson, we are going to explore the geometrical proof of identities.

Let us first see the varies identities which we will discuss here:

1. $\left(x+a\right)\left(x+b\right)\phantom{\rule{0.147em}{0ex}}={x}^{2}+x\left(a+b\right)+\mathit{ab}$

2. $$(a + b^2) = a^2 + 2ab + b^2$$

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

4.

Let us take each identity one by one and discuss the proof of that identity.

Identity $$1$$: $\left(x+a\right)\left(x+b\right)\phantom{\rule{0.147em}{0ex}}={x}^{2}+x\left(a+b\right)+\mathit{ab}$.

Let us construct a rectangular diagram of four regions. One region is square-shaped with a dimension of $$4 × 4$$ (Orange). Also, the other three regions are rectangle in shape with dimensions $$4×2$$ (Blue), $$3 × 4$$ (Yellow) and $$3 × 2$$ (Green). By observing the above rectangle, we can notice that:

$$\text{Area of the bigger rectangle}$$ $$=$$ $$\text{Area of a square (Orange)}$$ $$+$$ $$\text{Area of three rectangles}$$

$$(4 + 3)$$ $$(4 + 2)$$ $$=$$ $$(4 × 4) + (4 × 2) + (3 × 4) + (3 × 2)$$

Now, we simplify the LHS and RHS of the above expression.

LHS $$=$$ $$(4 + 3)$$ $$(4 + 2)$$ $$=$$ $$7×6 = 42$$

RHS $$=$$ $$(4 × 4) + (4 × 2) + (3 × 4) + (3 × 2)$$

RHS $$=$$ $$16+8+12+6 = 42$$

Therefore, LHS $$=$$ RHS

Similarly, if we use the variables in this case instead of number we get: Let one side of a rectangle be $$(x +a)$$, and the other side be $$(x + b)$$ units.

Then, $$\text{the total area of the rectangle }AEGI$$ $$=$$ $$\text{length } \times \text{breadth}$$ $$=$$ $$(x+a)(x+b)$$………….$$(1)$$

$$\text{The area of the rectangle }AEGI$$ $$=$$ $$\text{The area of the square }ABCD$$ $$+$$ $$\text{The area of the rectangle }BEFD$$ $$+$$ $$\text{The area of the rectangle }DFGH$$ $$+$$ $$\text{The area of the rectangle }CDHI$$.

$$\text{The area of the rectangle AEGI}$$ $$=$$ ${x}^{2}+\mathit{ax}+\mathit{ab}+\mathit{bx}$

$={x}^{2}+x\left(a+b\right)+\mathit{ab}$……………..$$(2)$$

From the equation, $$(3)$$ and $$(4)$$ we get $\left(x+a\right)\left(x+b\right)\phantom{\rule{0.147em}{0ex}}={x}^{2}+x\left(a+b\right)+\mathit{ab}$.

Therefore, $\left(x+a\right)\left(x+b\right)\phantom{\rule{0.147em}{0ex}}={x}^{2}+x\left(a+b\right)+\mathit{ab}$ is a identity.
Example:
Simplify the expression $\left(a+3\right)\left(a+11\right)$ using the identity $\left(x+a\right)\left(x+b\right)\phantom{\rule{0.147em}{0ex}}={x}^{2}+x\left(a+b\right)+\mathit{ab}$.

The expression is $\left(a+3\right)\left(a+11\right)$.

Now write the given expression $\left(a+3\right)\left(a+11\right)$ with respect to the given identity $$(x+a)(x+b)$$ $$=$$ $$x^2+(a+b)x+ab$$.

$\begin{array}{l}\left(x+a\right)\left(x+b\right)=\left(a+3\right)\left(a+11\right)\\ \\ =\left({a}^{2}+a\left(3+11\right)+3×11\right)\end{array}$

Now simplify the expression.

$\left({a}^{2}+14a+33\right)$.

Therefore, $\left(a+3\right)\left(a+11\right)$ $$=$$ $\left({a}^{2}+14a+33\right)$.