If x+frac{1}{x}=a and x^2+frac{1}{x^3}=b,then | vedclass

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(A) Given,$x+\frac{1}{x}=a$ and $x^2+\frac{1}{x^3}=b$. Squaring $x+\frac{1}{x}=a$,we get: $x^2+\frac{1}{x^2}+2=a^2 \Rightarrow x^2+\frac{1}{x^2}=a^2-2

Vedclass · Accepted Answer

$a^3+a^2-3a-2-b$

Vedclass · Answer

(A) Given,$x+\frac{1}{x}=a$ and $x^2+\frac{1}{x^3}=b$. Squaring $x+\frac{1}{x}=a$,we get: $x^2+\frac{1}{x^2}+2=a^2 \Rightarrow x^2+\frac{1}{x^2}=a^2-2$. Cubing $x+\frac{1}{x}=a$,we get: $x^3+\frac{1}{x^3}+3(x+\frac{1}{x})=a^3 \Rightarrow x^3+\frac{1}{x^3}=a^3-3a$. Now,consider the sum: $(x^2+\frac{1}{x^2}) + (x^3+\frac{1}{x^3}) = (a^2-2) + (a^3-3a)$. Rearranging the terms: $(x^2+\frac{1}{x^3}) + (x^3+\frac{1}{x^2}) = a^3+a^2-3a-2$. Substituting $x^2+\frac{1}{x^3}=b$: $b + (x^3+\frac{1}{x^2}) = a^3+a^2-3a-2$. Therefore,$x^3+\frac{1}{x^2} = a^3+a^2-3a-2-b$.

If $x+\frac{1}{x}=a$ and $x^2+\frac{1}{x^3}=b$,then $x^3+\frac{1}{x^2}$ is

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