Question
If $\text{x}-\frac{1}{\text{x}}=7,$ find the value of $\text{x}^3-\frac{1}{\text{x}^3}.$
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Answer
Given,If $\text{x}-\frac{1}{\text{x}}=7$We know that, $(a - b)^3 = a^3 - b^3 + 3ab(a + b) ...(1)$
Substitute $\text{x}-\frac{1}{\text{x}}=7$ in eq$(1)$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3=\text{x}^3-\frac{1}{\text{x}^3}-3\Big(\text{x}\times\frac{1}{\text{x}}\Big)\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\Rightarrow7^3=\text{x}^3-\frac{1}{\text{x}^3}-3\Big(\text{x}\times\frac{1}{\text{x}}\Big)\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\Rightarrow343=\text{x}^3-\frac{1}{\text{x}^3}-(3\times7)$
$\Rightarrow343=\text{x}^3-\frac{1}{\text{x}^3}-21$
$\Rightarrow125+21=\text{x}^3-\frac{1}{\text{x}^3}$
$\Rightarrow\text{x}^3-\frac{1}{\text{x}^3}=364$
Hence, the result is $\text{x}^3-\frac{1}{\text{x}^3}=364.$
Substitute $\text{x}-\frac{1}{\text{x}}=7$ in eq$(1)$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3=\text{x}^3-\frac{1}{\text{x}^3}-3\Big(\text{x}\times\frac{1}{\text{x}}\Big)\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\Rightarrow7^3=\text{x}^3-\frac{1}{\text{x}^3}-3\Big(\text{x}\times\frac{1}{\text{x}}\Big)\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\Rightarrow343=\text{x}^3-\frac{1}{\text{x}^3}-(3\times7)$
$\Rightarrow343=\text{x}^3-\frac{1}{\text{x}^3}-21$
$\Rightarrow125+21=\text{x}^3-\frac{1}{\text{x}^3}$
$\Rightarrow\text{x}^3-\frac{1}{\text{x}^3}=364$
Hence, the result is $\text{x}^3-\frac{1}{\text{x}^3}=364.$
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