+349 The last one was quite easy, but for the benefit of the people who didn't get it, here's the solution:
\(a+b=10, a^2+b^2=44,a^3+b^3=?\)
\((a+b)^2=10^2\)
\(a^2+b^2+2ab=100\)
\(44+2ab=100\)
\(2ab=56,ab=28\)
\(a^3+b^3=(a+b)(a^2-ab+b^2)\)
\(a^3+b^3=(10)(44-28)\)
\(a^3+b^3=10*16\)
\(a^3+b^3=160\)
For those who solved it, good job! Now for the next one:
\(wxy=10\)
\(wyz=5\)
\(wxz=45\)
\(xyz=12\)
Find \(w+x+y+z\)
GOOD LUCK!
Divide and substitute and you should get the following:
w = 2 1/2, x = 6, y = 2/3, z = 3, so have:
w+x+y+z = 2 1/2 + 6 + 2/3 + 3 =12 1/6
+26393 Now for the next one:
\(\begin{array}{rcr} wxy&=&10 \\ wyz&=&5 \\ wxz&=&45 \\ xyz&=&12 \\ \end{array} \)
Find \( w+x+y+z\)
1.
\(\begin{array}{|rcll|} \hline xyz &=& 12 \\ xyz \cdot \frac{wyz}{wxy} \cdot \frac{wxz}{wxy} &=& 12 \cdot \frac{5}{10} \cdot \frac{45}{10} \\ xyz \cdot \frac{z}{x} \cdot \frac{z}{y} &=& 12 \cdot \frac{5}{10} \cdot \frac{45}{10} \\ z^3 &=& 12 \cdot \frac{1}{2} \cdot \frac{9}{2} \\ z^3 &=& 3\cdot 9 \\ z^3 &=& 3^3 \\ \mathbf{z} & \mathbf{=} & \mathbf{3} \\ \hline \end{array} \)
2.
\(\begin{array}{|rcll|} \hline && w+x+y+z \\ &=& \left( \frac{wxy}{xyz} + \frac{wxy}{wyz} + \frac{wxy}{wxz} \right) \cdot z + z \\ &=& \left( \frac{10}{12} + \frac{10}{5} + \frac{10}{45} \right)\cdot z + z \\ &=& \left( \frac{5}{6} + 2 + \frac{2}{9} \right)\cdot z + z \quad & | \quad z= 3 \\ &=& \left( \frac{5}{6} + 2 + \frac{2}{9} \right)\cdot 3 + 3 \\ &=& \frac{5}{2} + 6 + \frac{2}{3} + 3 \\ &=& 9 + \frac{5}{2} + \frac{2}{3} \\ &=& 9 + \frac{15+4}{6} \\ &=& 9 + \frac{19}{6} \\ &=& 12 + \frac{1}{6} \\ \hline \end{array}\)
