Question
If $\text{x}+\frac{1}{\text{x}}=5,$ find the value of $\text{x}^3+\frac{1}{\text{x}^3}.$
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Answer
Given, $\text{x}+\frac{1}{\text{x}}=5$ We know that, $(a + b)^3 = a^3 + b^3 + 3ab(a + b) ...(1)$
Substitute $\text{x}+\frac{1}{\text{x}}=5$ in eq$(1)$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=\text{x}^3+\frac{1}{\text{x}^3}+3\Big(\text{x}\times\frac{1}{\text{x}}\Big)\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow5^3=\text{x}^3+\frac{1}{\text{x}^3}+3\Big(\text{x}\times\frac{1}{\text{x}}\Big)\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow125=\text{x}^3+\frac{1}{\text{x}^3}+3\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow125=\text{x}^3+\frac{1}{\text{x}^3}+3(5)$
$\Rightarrow125=\text{x}^3+\frac{1}{\text{x}^3}+15$
$\Rightarrow125-15=\text{x}^3+\frac{1}{\text{x}^3}$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=110$
Hence, the result is $\text{x}^3+\frac{1}{\text{x}^3}=110.$
Substitute $\text{x}+\frac{1}{\text{x}}=5$ in eq$(1)$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=\text{x}^3+\frac{1}{\text{x}^3}+3\Big(\text{x}\times\frac{1}{\text{x}}\Big)\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow5^3=\text{x}^3+\frac{1}{\text{x}^3}+3\Big(\text{x}\times\frac{1}{\text{x}}\Big)\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow125=\text{x}^3+\frac{1}{\text{x}^3}+3\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow125=\text{x}^3+\frac{1}{\text{x}^3}+3(5)$
$\Rightarrow125=\text{x}^3+\frac{1}{\text{x}^3}+15$
$\Rightarrow125-15=\text{x}^3+\frac{1}{\text{x}^3}$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=110$
Hence, the result is $\text{x}^3+\frac{1}{\text{x}^3}=110.$
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