If $$x^2-2x+1=$$​0 and x>0, then what is the value of $$x^2+x^3+\frac{1}{x^2} +\frac{1}{x^3} ?$$​​

Correct option is D

​Given:

$$x^2-2x+1=0 \\$$ 

Solution:

​$$x^2-2x+1=0 \\$$​​

Divide by x

​$$x-2+\frac1x = 0 \\x+ \frac 1x = 2$$ ........(1)

Squaring both sides equation 1.

​$$(x+\frac1x)^2 = 2^2\\x^2+ \frac{1}{x^2} +2 \times x\times\frac1x = 4\\x^2+ \frac{1}{x^2} = 4-2 = 2$$\

Cube the equation 1 both sides

​$$\begin{aligned}&\left(x+ \frac{1}{x}\right)^3 = 2^3 \\&x^3+\frac{1}{x^3}+ 3x \cdot \frac{1}{x} \cdot \left(x+\frac{1}{x}\right) = 8 \\&x^3+\frac{1}{x^3}+ 3(2)= 8 \quad \text{[From equation 1]} \\&x^3+\frac{1}{x^3} = 8-6 = 2\end{aligned}$$

​$$​​x^2+x^3+\frac{1}{x^2} +\frac{1}{x^3} \\​​x^2+\frac{1}{x^2}+x^3 +\frac{1}{x^3} \\2+2 = 4$$ 

Alternate Method:

When  $$x+ \frac 1x = 2$$ , There is always a fix value of x i.e. 1

​$$​​x^2+\frac{1}{x^2}+x^3 +\frac{1}{x^3} \\\ \\=​​1^2+\frac{1}{1^2}+1^3 +\frac{1}{1^3} \\=4$$​​

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