If a+b+c = 1, a²+b²+c² = 2 and a³+b³+c³ = 3, then find the value of a⁵+b⁵+c⁵ = ?

Step-by-step explanation:

Given :   a + b + c = 1

a² + b² + c² = 2

a³ + b³ + c³ = 3

To Find : a⁵ + b⁵ + c⁵

a² + b² + c² = 2

a³ + b³ + c³ = 3

Multiplying both

a⁵ + b⁵ + c⁵  + a²( b³ + c³) + b²(a³   + c³)  + c²(a³ + b³ )  = 6

=> a⁵ + b⁵ + c⁵ + a²b² (a + b)  +a²c²(a + c) + b²c²(b + c)  = 6

=>  a⁵ + b⁵ + c⁵ + a²b² (1-c)  +a²c²(1-b) + b²c²(1-a)  = 6

=> a⁵ + b⁵ + c⁵ + a²b² +a²c² + b²c²  - abc(ab + ac + bc)   = 6   Eq1

a + b + c = 1

squaring both sides

=> a² + b² + c²   + 2(ab + ac + bc) = 1

=> 2 + 2(ab + ac + bc) = 1

=> ab + ac + bc = - 1/2

Squaring both sides

=> a²b² +a²c² + b²c²  + 2abc(a +  b + c)  = 1/4

=> a²b² +a²c² + b²c²   + 2abc = 1/4

=> a²b² +a²c² + b²c² = 1/4  -  2abc

a⁵ + b⁵ + c⁵ + a²b² +a²c² + b²c²  - abc(ab + ac + bc)   = 6  

=> a⁵ + b⁵ + c⁵ +  1/4  -  2abc  - abc (-1/2)  = 6

=>  a⁵ + b⁵ + c⁵ = 23/4  + 3abc/2

a + b + c = 1

a² + b² + c² = 2

Multiplying both

a³ + b³ + c³  + a(b² + c² ) + b(a²   + c²) + c(a² + b²)  = 2

=> 3 +  ab (a + b) + ac(a + c)  +  bc(b + c) = 2

=>  ab(1 - c) + ac( 1 - b)  + bc( 1 - a)  = - 1

=> ab + ac + bc  -3abc  = - 1

ab + ac + bc = - 1/2

=> - 1/2 - 3abc = - 1

=> 3abc = 1/2

=> abc = 1/6

a⁵ + b⁵ + c⁵ = 23/4  + 3abc/2

=>  a⁵ + b⁵ + c⁵ = 23/4  + 1/4

=> a⁵ + b⁵ + c⁵ =  6

Hence, the value of a⁵ + b⁵ + c⁵ is  6.

answer: 1 2 3 6

step-by-step explanation:

1x6=6 2x3=6

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