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Given $$x + y = 5 \qquad xy = 1$$ Find $x^3 + y^3$.

To solve this, I tried this:

$y = \frac{1} {x}$

$x + \frac{1}{x} = 5$

$x^2 - 5x + 1 = 0$

What formula needs to be used to find the value of $x$?

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  • $\begingroup$ Here's a question. What happens when you expand out $(x-2.5)^2$? (This is not necessary for solving the problem.) $\endgroup$ Jan 30, 2023 at 1:25

5 Answers 5

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Another method that requires some intuition and knowledge of cubics:

Observe that, $$\large{\displaystyle{(\color{red}{x + y})^3 = \color{purple}{x}^3 + 3x^2y + 3xy^2 + \color{purple}{y}^3} = \color{purple}{x}^3 + 3\color{green}{xy}\big(\color{red}{x+y}\big) + \color{purple}{y}^3}$$

Otherwise re-arranged, gives:

$$\large{\displaystyle{\color{purple}{x}^3 + \color{purple}{y}^3 =(\color{red}{x + y})^3 - 3\color{green}{xy}\big(\color{red}{x+y}\big)}}$$

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$$ \begin{aligned} x^3+y^3 & =(x+y)^3-3 x y(x+y) \\ & =5^3-3(1)(5) \\ & =110 \end{aligned} $$

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  • $\begingroup$ On the money, neat. $\endgroup$
    – orangeskid
    Jan 30, 2023 at 8:11
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You can use the quadratic formula, or use a trick. Recall that $x^3+y^3=(x+y)(x^2-xy+y^2) $. Now consider that $x^2-xy+y^2=(x+y)^2-3xy$

So you have: $$x^3+y^3=(x+y)(x^2-xy+y^2)=(x+y)((x+y)^2-3xy) $$

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Note that $x^3+y^3=(x+y)(x^2-xy+y^3)$. We have $x+y=5$. Then $(x+y)^2=x^2+y^2+2xy=25$. Then, $x^2-xy+y^2=25-3xy=22$. All that is left is to multiply them.

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Use the binomial formula. We have $$(x+y)^3=x^3 + 3x^2y + 3xy^2 + y^3,$$ which reduces to $$125=x^3 + 3\cdot 5 + y^3$$ and thus $$x^3 + y^3 = 110.$$

There is no need to solve for $x$.

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