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# clever algebraic manipulations

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If $a+b=7$ and $a^3+b^3=42$, what is the value of the sum $\dfrac{1}{a}+\dfrac{1}{b}$? Express your answer as a common fraction.

Mar 24, 2021

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Solve for a and b: a = -(sqrt(87)*i - 21)/6 and b = (sqrt(87)*i + 21)/6 ==> 1/a + 1/b = 21/44

Mar 24, 2021
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Unfortunately, that's incorrect, and instead of actually finding the solutions, there is a way that requires less computation:

First, notice that $$\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}$$. Since we already know what a+b is, we just need to compute ab.

To do that, first factor the sum of cubes:

$$(a+b)(a^2-ab+b^2)=42$$

Then, replace a+b with 7:

$$a^2-ab+b^2=6$$

Then, square both sides of the first equation:

$$a^2+2ab+b^2=49$$

Then, subtract the first equation from the second one:

$$3ab=43\\ ab=\frac{43}{3}$$

Then, just replace ab with 43/3 and a+b with 7:

$$\frac{7}{\frac{43}{3}} = \frac{21}{43}$$

textot  Mar 24, 2021