If x2-3x+1=0 the then value of x3+1/x3 isIf x2-3x+1=0 the then value of x3+1/x3 is
D Debasish New member Joined Jan 19, 2012 Messages 13 Jan 20, 2012 #1 If x2-3x+1=0 the then value of x3+1/x3 isIf x2-3x+1=0 the then value of x3+1/x3 is
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Jan 20, 2012 #2 Hello, Debasish! This requires some Olympic-level gymnastics . . . \(\displaystyle \text{If }x^2-3x+1\,=\,0,\,\text{ find the value of }x^3+\dfrac{1}{x^3}\) Click to expand... \(\displaystyle \text{We are given: }\:x^2 + 1 \:=\:3x\) \(\displaystyle \text{Divide by }x\!:\;\;x + \dfrac{1}{x} \:=\:3\) \(\displaystyle \text{Cube: }\;\left(x + \dfrac{1}{x}\right)^3 \:=\:3^3 \quad\Rightarrow\quad x^3 + 3x + \dfrac{3}{x} + \dfrac{1}{x^3} \:=\:27 \) . . . . . . \(\displaystyle x^3 + \dfrac{1}{x^3} + 3\underbrace{\left(x + \dfrac{1}{x}\right)}_{\text{This is 3}} \:=\:27 \quad\Rightarrow\quad x^3 + \dfrac{1}{x^3} + 9 \:=\:27 \) \(\displaystyle \text{Therefore: }\;x^3 + \dfrac{1}{x^3} \:=\:18\)
Hello, Debasish! This requires some Olympic-level gymnastics . . . \(\displaystyle \text{If }x^2-3x+1\,=\,0,\,\text{ find the value of }x^3+\dfrac{1}{x^3}\) Click to expand... \(\displaystyle \text{We are given: }\:x^2 + 1 \:=\:3x\) \(\displaystyle \text{Divide by }x\!:\;\;x + \dfrac{1}{x} \:=\:3\) \(\displaystyle \text{Cube: }\;\left(x + \dfrac{1}{x}\right)^3 \:=\:3^3 \quad\Rightarrow\quad x^3 + 3x + \dfrac{3}{x} + \dfrac{1}{x^3} \:=\:27 \) . . . . . . \(\displaystyle x^3 + \dfrac{1}{x^3} + 3\underbrace{\left(x + \dfrac{1}{x}\right)}_{\text{This is 3}} \:=\:27 \quad\Rightarrow\quad x^3 + \dfrac{1}{x^3} + 9 \:=\:27 \) \(\displaystyle \text{Therefore: }\;x^3 + \dfrac{1}{x^3} \:=\:18\)
