Is it simple?

Algebra Level 3

Given $x, y, z$ such that

$\begin{array}{l r l} {\rm{If}} & x + y + z & = 1,\\ \quad & {x^2} + {y^2} + {z^2} & = 2,\\ \quad & {x^3} + {y^3} + {z^3} & = 3,\\ \end{array}$

what is the value of

$x^5 + y^5 + z^5$

to 2 decimal places?

The answer is 6.

1 solution

Mas Mus
Apr 20, 2014

Let

${E_1} = x + y + z = 1$

${E_2} = {x^2} + {y^2} + {z^2} = 2$

${E_3} = {x^3} + {y^3} + {z^3} = 3$

By manipulating ${E_1},{E_2}$ and ${E_3}$ we will find that ${E_4} = xy + xz + yz = - \frac{1}{2}$ and ${E_5} = xyz = \frac{1}{6}$

Now, evaluate ${E_6} = {x^4} + {y^4} + {z^4}$

$\begin{aligned} x^4 + y^4 + z^4 &= {\left( {{x^2} + {y^2} + {z^2}} \right)^2} - 2{\left( {xy + xz + yz} \right)^2} + 4\left( {xyz} \right)\left( {x + y + z} \right)\\&= 4 - 2\left( {\frac{1}{4}} \right) + 4\left( {\frac{1}{6}} \right)(1) = \frac{{25}}{6}\end{aligned}$

The last, evaluate ${E_7} = {x^5} + {y^5} + {z^5}$

$\begin{aligned} x^5 + y^5 + z^5 &= {\left( {x^4} + {y^4} + {z^4} \right)}\left( {x + y + z} \right) - \left({x^3}+{y^3}+{z^3}\right)\left( {xy + xz + yz}\right) + \left( {{x^2} + {y^2} + {z^2}} \right)\left( {xyz} \right)\\& = \left( {\frac{{25}}{6}} \right)(1) - (3)\left( { - \frac{1}{2}} \right) +(2) \left( {\frac{1}{6}} \right) = \frac{36}{6} =6\end{aligned}$

Should we simplify 33/6?

Kenny Lau - 6 years, 11 months ago

Of course. $\frac{33}{6}=\frac{11}{2}$ . Not only do I and Wolfram Alpha get a different answer, but also the $a,b$ in the problem can be anything you want since it is not given that $\gcd(a,b)=1$ . It is very surprising how anyone got this question right.

mathh mathh - 6 years, 10 months ago

Mas, the error in your approach is in calculating $x^5 + y^5 + z^5$ , which you applied the wrong Newton Identity. Instead, it should be

$x^5 + y^5 + z^5 = ( x^4 + y^4 + z^4) ( x+y+z) - (x^3 + y^3 +z^3 ) ( xy+yz+zx) + (x^2+y^2+z^2 ) ( xyz ) .$

This gives us the value of 6.

Those who previously answered 7 have been marked correct. I have updated the phrasing of the question and the answer is now 6.

Calvin Lin Staff - 6 years, 5 months ago

In response to @Calvin Lin thank you for the correction

Mas Mus - 6 years, 5 months ago

Can you update your solution accordingly? Thanks!

Calvin Lin Staff - 6 years, 5 months ago

@Calvin Lin – In response to @Calvin Lin : I have updated the solution. Thanks

Mas Mus - 6 years, 2 months ago

Answer should be 7. I think.

Mayur Choudhary - 7 years, 1 month ago

How do you find the value of xyz?

Kenny Lau - 6 years, 11 months ago

According to Wolfram ALpha , the answer is 6.

Kenny Lau - 6 years, 11 months ago

I got exactly 6 after using Newton sums. Create a polynomial $f(a)=a^3-a^2-0.5a-\frac{1}{6}=0$ . The roots ( $x,y,z$ ) of this polynomial satisfy everything that is given and after using Newton sums I get that the roots also satisfy $x^5+y^5+z^5=6$ .

I wrote my answer as $7$ , since $6=\frac{6}{1}$ .

mathh mathh - 6 years, 10 months ago

a , b could be any number provided the fractional value is correct . Hence a+b= 39 is one of the answers.

Bhunit Santhiramoulesan - 7 years, 1 month ago

Is it simple?

The answer is 6.

1 solution

0 pending reports