Question
Simplify the following products:$\Big(\text{m}+\frac{\text{n}}{7}\Big)^3\Big(\text{m}-\frac{\text{n}}{7}\Big)$
✓
Answer
In the given problem, we have to find product of $\Big(\text{m}+\frac{\text{n}}{7}\Big)^3\Big(\text{m}-\frac{\text{n}}{7}\Big)$ We have been given $\Big(\text{m}+\frac{\text{n}}{7}\Big)^3\Big(\text{m}-\frac{\text{n}}{7}\Big)$ On rearranging we get,$\Big(\text{m}+\frac{\text{n}}{7}\Big)^3\Big(\text{m}-\frac{\text{n}}{7}\Big)=\Big(\text{m}+\frac{\text{n}}{7}\Big)^2\Big(\text{m}+\frac{\text{n}}{7}\Big)\Big(\text{m}-\frac{\text{n}}{7}\Big)$
We shall use the identity $\big(\text{x}-\text{y}\big)\big(\text{x}+\text{y}\big)=\text{x}^2-\text{y}^2$ By substituting $\text{x}=\text{m},\ \text{y}=\frac{\text{n}}{7},$ we get,$\Big(\text{m}+\frac{\text{n}}{7}\Big)^3\Big(\text{m}-\frac{\text{n}}{7}\Big)$
$=\Big(\text{m}+\frac{\text{n}}{7}\Big)^2\bigg(\text{m}^2-\Big(\frac{\text{n}}{7}\Big)^2\bigg)$
$=\Big(\text{m}+\frac{\text{n}}{7}\Big)^2\Big(\text{m}^2-\frac{\text{n}^2}{49}\Big)$
Hence the value of $\Big(\text{m}+\frac{\text{n}}{7}\Big)^3\Big(\text{m}-\frac{\text{n}}{7}\Big)$ is $\Big(\text{m}+\frac{\text{n}}{7}\Big)^2\Big(\text{m}^2-\frac{\text{n}^2}{49}\Big)$
We shall use the identity $\big(\text{x}-\text{y}\big)\big(\text{x}+\text{y}\big)=\text{x}^2-\text{y}^2$ By substituting $\text{x}=\text{m},\ \text{y}=\frac{\text{n}}{7},$ we get,$\Big(\text{m}+\frac{\text{n}}{7}\Big)^3\Big(\text{m}-\frac{\text{n}}{7}\Big)$
$=\Big(\text{m}+\frac{\text{n}}{7}\Big)^2\bigg(\text{m}^2-\Big(\frac{\text{n}}{7}\Big)^2\bigg)$
$=\Big(\text{m}+\frac{\text{n}}{7}\Big)^2\Big(\text{m}^2-\frac{\text{n}^2}{49}\Big)$
Hence the value of $\Big(\text{m}+\frac{\text{n}}{7}\Big)^3\Big(\text{m}-\frac{\text{n}}{7}\Big)$ is $\Big(\text{m}+\frac{\text{n}}{7}\Big)^2\Big(\text{m}^2-\frac{\text{n}^2}{49}\Big)$
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