Find the value of M

**Dragon** · Jan 30th 2007, 02:38 PM

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

**ThePerfectHacker** · Jan 30th 2007, 03:55 PM

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

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

**Soroban** · Jan 30th 2007, 08:14 PM

Hello, Dragon!

Another approach . . .

If $m+n\,=\,3$ and $m^2+n^2\,=\,6$ ,
find the numerical value for $m^3+n^3$

We are given: . $\begin{array}{cc}(1)\\(2)\end{array} \begin{array}{cc}m + n \:=\:3 \\ m^2+n^2\:=\:6\end{array}$

Square (1): . $(m + n)^2\:=\:3^2$

. . . . . . . . . $\underbrace{m^2 + n^2} + 2mn\:=\:9$
. . . . . . . . . - - $\downarrow$
Substitute (2): . $6 + 2mn \:=\:9\quad\Rightarrow\quad mn \,=\,\frac{3}{2}\quad(3)$

Cube (1): . $(m+n)^3\:=\:3^3\quad\Rightarrow\quad m^3 + 3m^2n + 3mn^2 + n^3\:=\:27$

. . . . . . . $\text{and we have: }\;\;m^3 + n^3 + 3\underbrace{mn}\underbrace{(m + n)}\:=\:27$
. . . . . . . . . . . . . . . . . . . . . . . . . . $\downarrow\quad\;\downarrow$
Substitute (3) and (1): . $m^3 + n^3 + 3\left(\frac{3}{2}\right)(3)\:=\:27$

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

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