Practicing Success
If $x+y=4$ and $\frac{1}{x}+\frac{1}{y}=\frac{16}{15}$, then what is the value of $\left(x^3+y^3\right) ?$ |
18 16 19 21 |
19 |
If x + y = n then, $x^3 + y^3$ = n3 - 3 × n × xy $x+y=4$ $\frac{1}{x}+\frac{1}{y}=\frac{16}{15}$, Then what is the value of $\left(x^3+y^3\right) ?$ $\frac{1}{x}+\frac{1}{y}=\frac{16}{15}$, \(\frac{x + y}{xy}\) = $\frac{16}{15}$ \(\frac{4}{xy}\) = $\frac{16}{15}$ xy = $\frac{15}{4}$ Then, $\left(x^3+y^3\right) $ = 43 - 3 × 4 × $\frac{15}{4}$] $\left(x^3+y^3\right) $ = 64 - 45 = 19 |