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.

ellapow Mar 24, 2021

#1**0 **

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

Guest Mar 24, 2021

#2**+1 **

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