Question
If $\text{x}^2+\frac{1}{\text{x}^2}=98,$ find the value of $\text{x}^3+\frac{1}{\text{x}^3}.$
✓
Answer
Given, $\text{x}^2+\frac{1}{\text{x}^2}=98$
We know that, $(x + y)^2 = x^2 + y^2 + 2xy ...(1)$
Substitute $\text{x}^2+\frac{1}{\text{x}^2}=98$ in eq.(1)
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2\times\text{x}\times\frac{1}{\text{x}}$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=98+2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=100$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\sqrt{100}$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\pm10$
We need to find $\text{x}^3+\frac{1}{\text{x}^3}$
So, $a^3 + b^3 = (a - b)(a^2 + b^2 - ab)$
$\text{x}^3+\frac{1}{\text{x}^3}=\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)-\Big(\text{x}\times\frac{1}{\text{x}}\Big)$
We know that,
$\Big(\text{x}+\frac{1}{\text{x}}\Big)=10$ and $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=98$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=10(98-1)$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=10(97)$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=970$
Hence, the value of $\text{x}^3+\frac{1}{\text{x}^3}=970.$
We know that, $(x + y)^2 = x^2 + y^2 + 2xy ...(1)$
Substitute $\text{x}^2+\frac{1}{\text{x}^2}=98$ in eq.(1)
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2\times\text{x}\times\frac{1}{\text{x}}$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=98+2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=100$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\sqrt{100}$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\pm10$
We need to find $\text{x}^3+\frac{1}{\text{x}^3}$
So, $a^3 + b^3 = (a - b)(a^2 + b^2 - ab)$
$\text{x}^3+\frac{1}{\text{x}^3}=\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)-\Big(\text{x}\times\frac{1}{\text{x}}\Big)$
We know that,
$\Big(\text{x}+\frac{1}{\text{x}}\Big)=10$ and $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=98$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=10(98-1)$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=10(97)$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=970$
Hence, the value of $\text{x}^3+\frac{1}{\text{x}^3}=970.$
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