If m+n=3 and m^2+N^2=6 find the numerical value for m^3+N^3
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If $\displaystyle m+n\,=\,3$ and $\displaystyle m^2+n^2\,=\,6$,
find the numerical value for $\displaystyle m^3+n^3$
We are given: .$\displaystyle \begin{array}{cc}(1)\\(2)\end{array} \begin{array}{cc}m + n \:=\:3 \\ m^2+n^2\:=\:6\end{array}$
Square (1): .$\displaystyle (m + n)^2\:=\:3^2$
. . . . . . . . . $\displaystyle \underbrace{m^2 + n^2} + 2mn\:=\:9$
. . . . . . . . . - - $\displaystyle \downarrow$
Substitute (2): .$\displaystyle 6 + 2mn \:=\:9\quad\Rightarrow\quad mn \,=\,\frac{3}{2}\quad(3)$
Cube (1): .$\displaystyle (m+n)^3\:=\:3^3\quad\Rightarrow\quad m^3 + 3m^2n + 3mn^2 + n^3\:=\:27$
. . . . . . . $\displaystyle \text{and we have: }\;\;m^3 + n^3 + 3\underbrace{mn}\underbrace{(m + n)}\:=\:27$
. . . . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle \downarrow\quad\;\downarrow$
Substitute (3) and (1): .$\displaystyle m^3 + n^3 + 3\left(\frac{3}{2}\right)(3)\:=\:27$
Therefore: .$\displaystyle m^3+n^3\:=\:\frac{27}{2}$