Math, asked by Anonymous, 1 month ago

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]​

Attachments:

Answers

Answered by naveenprasad10a
11

Step-by-step explanation:

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

Answered by IamIronMan0
18

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

Similar questions