Question

If
${x}^{2} + \frac{1}{ {x}^{2} } = 98$
Find the value of
${x}^{3} + \frac{1}{ {x}^{3} } =$
​

Answer 1

$\huge\mathbb{SOLUTION:-}$

$\bold{Given}\begin{cases}\sf{x{}^{2}+\frac{1}{x{}^{2}}=98}\end{cases}$

$\huge\sf{To\:Find}$

$\sf x{}^{3}+\frac{1}{x{}^{3}}$

Now :-

$\bf\underline{\underline{\red{According\:To\: Question:-}}}$

• First find the value of
• $\sf x+\frac{1}{x}$

$\sf\longmapsto [x+\frac{1}{x}]{}^{2}=x+\frac{1}{x{}^{2}}+2\\ \\ \sf\longmapsto[x+\frac{1}{x}]{}^{2}=98+2\\ \\ \sf\longmapsto [x+\frac{1}{x}]{}^{2}=100\\ \\ \sf\longmapsto x+\frac{1}{x}=\sqrt{100}\\ \\ \sf\implies x+\frac{1}{x}=10$

$\boxed{\mathfrak{\purple{x+\frac{1}{x}=10}}}$

• Now we have to find the value of $\sf x{}^{3}+\frac{1}{x{}^{3}}$

$\sf\leadsto [x+\frac{1}{x}]{}^{3}=x{}^{3}+\frac{1}{x{}^{3}}+3\times \cancel{x}\times \frac{1}{\cancel{x}}(x+\frac{1}{x})\\ \\ \sf\leadsto [x+\frac{1}{x}]{}^{3}=x{}^{3}+\frac{1}{x{}^{3}}+3(x+\frac{1}{x})$

$\sf\implies (10){}^{3}=x{}^{3}+\frac{1}{x{}^{3}}+3\times 10\\ \\ \sf\implies 1000=x{}^{3}+\frac{1}{x{}^{3}}+30\\ \\ \sf\implies 1000-30=x{}^{3}+\frac{1}{x{}^{3}}\\ \\ \sf\implies 970=x{}^{3}+\frac{1}{x{}^{3}}$

$\boxed{\boxed{\mathfrak{\red{x{}^{3}+\frac{1}{x{}^{3}}=970}}}}$

#BAL

#answerwithquality