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If $x+\frac{1}{x}=a, x^{2}+\frac{1}{x^{3}}=b$, then $x^{3}+\frac{1}{x^{2}}$ is -
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Verified Answer
The correct answer is:
$a^{3}+a^{2}-3 a-2-b$
$x+\frac{1}{x}=a$ and $x^{2}+\frac{1}{x^{3}}=b$
$\left(x+\frac{1}{x}\right)^{2}=a^{2} \Rightarrow x^{2}+\frac{1}{x^{2}}+2=a^{2}$ ...(1)
and $\left(x+\frac{1}{x}\right)^{3}=a^{3} \Rightarrow x^{3}+\frac{1}{x^{3}}+3 x \cdot \frac{1}{x}\left(x+\frac{1}{x}\right)=a^{3}$ ...(2)
adding (1) \& (2)
$\begin{array}{l}
\left(x^{2}+\frac{1}{x^{3}}\right)+\left(x^{3}+\frac{1}{x^{2}}\right)+2+3\left(x+\frac{1}{x}\right)=a^{2}+a^{3} \\
b+\left(x^{3}+\frac{1}{x^{2}}\right)+2+3 a=a^{2}+a^{3} \\
\therefore x^{3}+\frac{1}{x^{2}}=a^{3}+a^{2}-3 a-2-b
\end{array}$
$\left(x+\frac{1}{x}\right)^{2}=a^{2} \Rightarrow x^{2}+\frac{1}{x^{2}}+2=a^{2}$ ...(1)
and $\left(x+\frac{1}{x}\right)^{3}=a^{3} \Rightarrow x^{3}+\frac{1}{x^{3}}+3 x \cdot \frac{1}{x}\left(x+\frac{1}{x}\right)=a^{3}$ ...(2)
adding (1) \& (2)
$\begin{array}{l}
\left(x^{2}+\frac{1}{x^{3}}\right)+\left(x^{3}+\frac{1}{x^{2}}\right)+2+3\left(x+\frac{1}{x}\right)=a^{2}+a^{3} \\
b+\left(x^{3}+\frac{1}{x^{2}}\right)+2+3 a=a^{2}+a^{3} \\
\therefore x^{3}+\frac{1}{x^{2}}=a^{3}+a^{2}-3 a-2-b
\end{array}$
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