Question

If the below three equations are true,
• x + y + z = 1
• x² + y² + z² = 2
• x³ + y³ + z³ = 3

Then find the value of x⁵ + y⁵ + z⁵

Note - If you are thinking '5' so no, it's not a correct solution.

Show your method along with explanation.

[Check the attachment for some hint]​

Answer 1

Step-by-step explanation:

x5+y5+z5 is 15 5+5+5=15

Answer 2

Answer:

$\huge \pink{ \sum {x}^{5} = 6}$

Step-by-step explanation:

We know

$(x + y + z) {}^{2} = \sum {x}^{2} + 2 \sum xy \\ \\ \implies \: 2 \sum xy = 1 - 2 =-1\\ \\ \implies \sum xy = - \frac{1}{2}$

Now a cubic equation whose root are x , y and z is given by

${t}^{3} - ( \sum x) {t}^{2} + ( \sum xy)t + c = 0 \\ \\ \implies {t}^{3} - {t}^{2} - \frac{1}{2} t + c = 0$

All three x , y and z are roots of this equation , satisfies them in equation and add all three , let's call this process taking summation of roots , we get

$\sum {x}^{3} - \sum {x {}^{2} }^{} - \frac{1}{2} \sum x +3 c = 0 \\ \\ \implies \: 3 - ( {2} ) - \frac{1}{2} (1) + 3c = 0 \\ \\ \implies \: c = \frac{ - 1}{6}$

Multiply Cubic equation by t , we get

${t}^{4} - {t}^{3} - \frac{1}{2} {t}^{2} - \frac{1}{6} t = 0$

Again take summation

$\sum {x}^{4} - \sum {x {}^{3} }^{} - \frac{1}{2} \sum {x}^{2} - \frac{1}{6} \sum x= 0 \\ \\ \implies \: \sum {x}^{4} = 3 + \frac{1}{2} \times 2 + \frac{1}{6} = \frac{25}{6}$

Again multiply equation by t

${t}^{5} - {t}^{4} - \frac{1}{2} {t}^{3} - \frac{1}{6} {t}^{2} = 0$

Take summation

$\sum {x}^{5} - \sum {x {}^{4} }^{} - \frac{1}{2} \sum x {}^{3} - \frac{ 1}{6} {x}^{2} = 0 \\ \\ \sum \: {x}^{5} = \frac{25}{6} + \frac{1}{2} (3) + \frac{1}{6} (2) \\ \\ \implies \sum {x}^{5} = 6$