How to use $x + \frac{1}{x} = 7$ to compute $x^2 + \frac{1}{x^2}$.

Notice $\left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2$.


Observe that
$$x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right )^2-2.$$
Hope this helps. :)


Thank you to Yiyuan Lee and awllower who really answered the question.

I am only adding this for completeness, and to remind myself of the solution.

We are given $x + \frac{1}{x} =7$ and want to use this to find $x^2 + \frac{1}{x^2}$.

Using the fact $$\left(x +\frac{1}{x}\right)^2 = x^2 +\frac{1}{x^2} +2$$ and rewriting this as: $$\left(x +\frac{1}{x}\right)^2 -2 = x^2 +\frac{1}{x^2}$$ We can substitute $x + \frac{1}{x} =7$ to get: $$(7)^2 -2 = x^2 +\frac{1}{x^2}$$ So $$x^2 +\frac{1}{x^2} = 47$$ The same method can be used to calculate $x^3 + \frac{1}{x^3}$: $$\left(x +\frac{1}{x}\right)^3 = x^3 +\frac{1}{x^3} +3x +\frac{3}{x}$$ Noticing that $3x +\frac{3}{x}$ = $3(x+ \frac{1}{x})$, which means $3x +\frac{3}{x} = 3.(7) = 21$

Then it is possible to write: $$\left(x +\frac{1}{x}\right)^3 -21 = x^3 + \frac{1}{x^3}$$ $$(7)^3 -21 = x^3 + \frac{1}{x^3}$$ $$x^3 + \frac{1}{x^3} = 343 -21$$ $$x^3 + \frac{1}{x^3} = 322$$

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Algebra Precalculus

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