Thread: Find the value of M

If m+n=3 and m^2+N^2=6 find the numerical value for m^3+N^3

Originally Posted by Dragon

If m+n=3 and m^2+N^2=6 find the numerical value for m^3+N^3

You are told that,
$\displaystyle n+m=3$
Thus, squaring
$\displaystyle (n+m)^2=9$
$\displaystyle n^2+2nm+m^2=9$
Thus,
$\displaystyle 2mn+6=9$
Thus,
$\displaystyle 2 mn=3 \,$
$\displaystyle mn=1.5$.

Thus,
$\displaystyle n^3+m^3=(n+m)(n^2-nm+m^2)=(3)(6-1.5)$

Hello, Dragon!

Another approach . . .

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: .$\displaystyle m^3+n^3\:=\:\frac{27}{2}$