Math, asked by bpjaiswal1944, 7 hours ago

a+b+c=4
a2+b2+c2=10
a3+b3+c3=22
a4+b4+c4= ?​

Answers

Answered by ruhanipasrija00
1

Answer:

Given :

a+b+c=4,

a²+b²+c²=10,

a³+b³+c³ = 22

To Find : a⁴+b⁴+c⁴=?

Solution:

a+b+c=4

Squaring both sides

=> a²+b²+c²  + 2(ab + bc + ca) = 16

=> 10 + 2(ab + bc + ca) = 16

=> ab + bc + ca = 3

a³+b³+c³  - 3abc  = (a + b + c)(a²+b²+c² - ( ab + bc + ca))

=> 22 - 3abc = (4)(10 - 3)

=> 22 - 3abc = 28

=> 3abc = - 6

=> abc = - 2

ab + bc + ca = 3

Squaring both sides

=> (ab)² + (bc)² + (ac)² + 2(ab.bc + ab.ca + bc.ca)  = 9

=>  (ab)² + (bc)² + (ac)²  + 2abc(a + b + c)  = 9

=> (ab)² + (bc)² + (ac)²  + 2(-2)(4)  = 9

=>  (ab)² + (bc)² + (ac)² = 25

=> a²b² + b²c² + a²c² = 25

a²+b²+c²=10

squaring both sides

=> a⁴ + b⁴ + c⁴  + 2( a²b² + b²c² + a²c²) = 100

=> a⁴ + b⁴ + c⁴  + 2( 25) = 100

=> a⁴ + b⁴ + c⁴  + 50 = 100

=>  a⁴ + b⁴ + c⁴ = 50

Step-by-step explanation:

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Answered by aaronharis30
1

Answer:

Given :

a+b+c=4,

a²+b²+c²=10,

a³+b³+c³ = 22

To Find : a⁴+b⁴+c⁴=?

Solution:

a+b+c=4

Squaring both sides

=> a²+b²+c²  + 2(ab + bc + ca) = 16

=> 10 + 2(ab + bc + ca) = 16

=> ab + bc + ca = 3

a³+b³+c³  - 3abc  = (a + b + c)(a²+b²+c² - ( ab + bc + ca))

=> 22 - 3abc = (4)(10 - 3)

=> 22 - 3abc = 28

=> 3abc = - 6

=> abc = - 2

ab + bc + ca = 3

Squaring both sides

=> (ab)² + (bc)² + (ac)² + 2(ab.bc + ab.ca + bc.ca)  = 9

=>  (ab)² + (bc)² + (ac)²  + 2abc(a + b + c)  = 9

=> (ab)² + (bc)² + (ac)²  + 2(-2)(4)  = 9

=>  (ab)² + (bc)² + (ac)² = 25

=> a²b² + b²c² + a²c² = 25

a²+b²+c²=10

squaring both sides

=> a⁴ + b⁴ + c⁴  + 2( a²b² + b²c² + a²c²) = 100

=> a⁴ + b⁴ + c⁴  + 2( 25) = 100

=> a⁴ + b⁴ + c⁴  + 50 = 100

=>  a⁴ + b⁴ + c⁴ = 50

Step-by-step explanation:

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