Question

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

• A. $a^3+a^2-3a-2-b$
• B. $a^3-a^2-3a+4-b$
• C. $a^3-a^2+3a-6-b$
• D. $a^3+a^2+3a-16-b$

Answer 1

The correct option is A $a^3+a^2-3a-2-b$

$x+\frac{1}{x}=a,x^2+\frac{1}{x^3}=b{(x+\frac{1}{x})}^2=a^2{x}^2+\frac{1}{x^2}+2=a^2.....\left(i\right){(x+\frac{1}{x})}^3=a^3{x}^3+\frac{1}{x^3}+3(x+\frac{1}{x})=a^3....\left(ii\right)$

Adding equation $\left(i\right)$ and $\left(i\right)$

$x^3+\frac{1}{x^3}+3(x+\frac{1}{x})+x^2+\frac{1}{x^2}+2=a^2+a^3{x}^2+\frac{1}{x^3}+x^3+\frac{1}{x^2}+3(x+\frac{1}{x})+2=a^2+a^{3}b+x^3+\frac{1}{x^2}+3a+2=a^2+a^3{x}^3+\frac{1}{x^2}=a^2+a^3-3a-b-2$

So option $A$ is correct