Question

Hey Guys!
Try to solve this maths question.....
$a + b + c = 4 \\ {a}^{2} + {b}^{2} + {c}^{2} = 10 \\ {a}^{3} + {b}^{3} + {c}^{3} = 22 \\ {a}^{4} + {b}^{4} + {c}^{4} = ?$
Don't copy from web and Don't spam
Need answer as well as explanation
Thank You!

​

Answer 1

Use the multinomial theorem to compute some polynomial expansions:

(a + b + c)² = a² + b² + c² + 2 (ab + ac + bc)

(a + b + c)³ = a³ + b³ + c³

… … … … … … + 3 (a²b + a²c + ab² + b²c + ac² + bc²)

… … … … … … + 6 abc

(a + b + c)⁴ = a⁴ + b⁴ + c⁴

… … … … … … + 4 (a³b + a³c + ab³ + b³c + ac³ + bc³)

… … … … … … + 6 (a²b² + a²c² + b²c²)

… … … … … … + 12 (a²bc + ab²c + abc²)

Now, given that

a + b + c = 4

a² + b² + c² = 10

it follows that

ab + ac + bc = (4² - 10)/2 = 3

We use this result to simplify the extra terms in the 3rd-degree expansion.

a²b + a²c + ab² + b²c + ac² + bc²

= a (ab + ac) + b (ab + bc) + c (ac + bc)

= a (ab + ac + bc) + b (ab + ac + bc) + c (ab + ac + bc) - 3abc

… … I underline the terms that are added and subtracted … …

= (a + b + c) (ab + ac + bc) - 3abc

= 4 • 3 - 3abc

= 12 - 3abc

Given that

a³ + b³ + c³ = 22

this tells us that

4³ = 22 + 3 (12 - 3abc) + 6abc

4³ = 58 - 3abc

and so

abc = (4³ - 58)/(-3) = -2

from which it follows that

a²b + a²c + ab² + b²c + ac² + bc² = 12 - 3 • (-2) = 18

Now we similarly simplify the extra terms in the 4th-degree expansion.

a³b + a³c + ab³ + b³c + ac³ + bc³

= a² (ab + ac) + b² (ab + bc) + c² (ac + bc)

= a² (ab + ac + bc) + b² (ab + ac + bc) + c² (ab + ac + bc) - a²bc - ab²c - abc²

= (a² + b² + c²) (ab + ac + bc) - (a²bc + ab²c + abc²)

= 10 • 3 - (a²bc + ab²c + abc²)

which means

4⁴ = a⁴ + b⁴ + c⁴

… … … … … … + 4 (30 - a²bc - ab²c - abc²)

… … … … … … + 6 (a²b² + a²c² + b²c²)

… … … … … … + 12 (a²bc + ab²c + abc²)

reduces to

136 = a⁴ + b⁴ + c⁴

… … … … … … + 6 (a²b² + a²c² + b²c²)

… … … … … … + 8 (a²bc + ab²c + abc²)

Now, we know a + b + c = 4 and abc = -2, so we have

a²bc + ab²c + abc² = abc (a + b + c) = -2 • 4 = -8

and so

136 = a⁴ + b⁴ + c⁴ + 6 (a²b² + a²c² + b²c²) + 8 • (-8)

200 = a⁴ + b⁴ + c⁴ + 6 (a²b² + a²c² + b²c²)

Finally, we know that ab + ac + bc = 3. Squaring this gives

(ab + ac + bc)² = 3²

a²b² + a²c² + b²c² + 2 (a²bc + ab²c + abc²) = 9

but we also know that a²bc + ab²c + abc² = -8, so

a²b² + a²c² + b²c² + 2 • (-8) = 9

a²b² + a²c² + b²c² = 25

Therefore, we end up with

200 = a⁴ + b⁴ + c⁴ + 6 • 25

⇒   [[[   a⁴ + b⁴ + c⁴ = 50   ]]