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# help algebra

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Find one ordered pair $(x,y)$ of real numbers such that $x + y = 10$ and $x^3 + y^3 = 162 + x^2 + y^2.$

Jun 10, 2024

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You might want to double check the question, because there is no real pair of numbers that holds those properties. There are, however, complex numbers that do:

$$y = 10 - x$$

$$x^3 + (10-x)^3 = 162 +x^2 + (10-x)^2$$

$$x^3 + (10-x)^2(10-x) = 162 + x^2 + (100 - 20x + x^2)$$

$$x^3 + (100-20x+x^2)(10-x) = 2x^2-20x+262$$

$$x^3 + (1000 -200x+10x^2-100x+20x^2-x^3) = 2x^2-20x+262$$

$$30x^2-300x+1000 = 2x^2 - 20x + 262$$

$$15x^2-150x+500=x^2-10x+131$$

$$14x^2-140x+369 = 0$$

$$x^2-10x+\frac{369}{14} = 0$$

$$x^2 - 10x +\frac{350}{14} = -\frac{19}{14}$$

$$x^2 - 10x + 25 = -\frac{19}{14}$$

$$(x-5)^2 = -\frac{19}{14}$$

$$x-5 = \sqrt{\frac{19}{14}}i$$

$$x = 5 + \sqrt{\frac{19}{14}}i$$

$$y = 10 - (5 + \sqrt{\frac{19}{14}}i)$$

$$y = 5 - \sqrt{\frac{19}{14}}i$$

Therefore, one ordered pair of complex numbers that works is 5 + sqrt(19/14)i and 5 - sqrt(19/14)i