Question

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$\sf \: a + b + c = 4 \\ \\ \sf \: {a}^{2} + {b}^{2} + {c}^{2} = 10 \\ \\ \sf \: {a}^{3} + {b}^{3} + {c}^{3} = 24 \\ \\ \sf \: then \: find \: {a}^{4} + {b}^{4} + {c}^{4}$
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Answer 1

Answer:

Given :

a+b+c=4,

a²+b²+c²=10,

a³+b³ + c³ = 22

To Find : a++b+c+=?

Solution:

a+b+c=4

Squaring both sides

=> a+b+c² + 2(ab + bc + ca) = 16

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

=> ab + bc + ca = 3

a³+b³ + c³ -3abc = (a + b + c)(a²+b²2+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

a²+b²+c²=10

squaring both sides

=> a+b+c+ + 2(a²b² + b²c² + a²c²) = 100

=> a+b+c² + 2(25) = 100

=> a+b+c4 +50= 100

=> a + b + c = 50

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

If a+b+c=6 and a2+b2+c2=14 and a3+b3+c3-36 find value of abc ...

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