Question

# ${\left(\frac{a}{4}-\frac{b}{2}+1\right)}^{2}$expand it with suitable identities .

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Solution

## Solve the given identity : Here we use the identity $\left({\text{x+y+z)}}^{2}{\text{=x}}^{2}{\text{+y}}^{2}{\text{+z}}^{2}\text{+2xy+2yz+2zx}$Now we have $\text{x=}\frac{\text{a}}{4}\text{,y=}\frac{\text{-b}}{2}\text{,z=1}$So ${\left(\frac{\text{a}}{4}-\frac{\text{b}}{2}+1\right)}^{2}=\left(\frac{\text{a}}{4}{\right)}^{2}+{\left(\frac{-\text{b}}{2}\right)}^{2}+{1}^{2}+2×\frac{\text{a}}{4}×\left(-\frac{\text{b}}{2}\right)+2\left(-\frac{\text{b}}{2}\right)×1+2×1×\left(\frac{\text{a}}{4}\right)$ $=\frac{{\text{a}}^{2}}{16}+\frac{{\text{b}}^{2}}{4}+1-\frac{\text{ab}}{4}-\text{b+}\frac{\text{a}}{2}$Hence ${\left(\frac{\text{a}}{4}-\frac{\text{b}}{2}+1\right)}^{2}$$=\frac{{\text{a}}^{2}}{16}+\frac{{\text{b}}^{2}}{4}+1-\frac{\text{ab}}{4}-\text{b+}\frac{\text{a}}{2}$

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