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Algebra Level 3

${ x }^{ 2\quad \quad }-3x+1\quad =\quad 0\\ then,\quad find\quad { x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } ?$

1 solution

Lawrence Bush
Aug 27, 2014

$x^{3}+\frac{1}{x^{3}}=(x+\frac{1}{x})^{3}-3*x*\frac{1}{x}(x+\frac{1}{x})$ .

$x^{3}+\frac{1}{x^{3}}=(x+\frac{1}{x})^{3}-3*(x+\frac{1}{x})$ . If $x+\frac{1}{x}=y$ ,then

$x^{3}+\frac{1}{x^{3}}=y^{3}-3y$ . Back to our equation.

Divide both sides by $x$ to get:

$x+\frac{1}{x}=3 \Rightarrow y=3$ . Easy we get $x^{3}+\frac{1}{x^{3}}=18$ .