Math problem: A+B+C=4 A^2+B^2+C^2= 10 A^3+B^3+C^3= 22 What is A^4+B^4+C^4= ?

Question

My friend has this problem for homework (algebra 2) She does not have a text book, so I do not know what concept this is covering. It looks like it should be easy, I must be looking at it incorrectly.

A+B+C=4

A^2+B^2+C^2= 10

A^3+B^3+C^3= 22

What is A^4+B^4+C^4= ?

If use 0 for A, 1 for B, and 3 for C I get 4, 10 and 28. Maybe she copied the problem incorrectly.

Answer 1

A + B + C = 4



(A + B + C)² = 16

A² + B² + C² + 2AB + 2BC + 2AC = 16

10 + 2AB + 2BC + 2AC = 16

2AB + 2BC + 2AC = 6

AB + BC + AC = 3

square both sides

A²B² + B²C² + A²C² + 2AB²C + 2ABC² + 2A²BC = 9

A²B² + B²C² + A²C² + 2ABC(A + B + C) = 9

A²B² + B²C² + A²C² + 2ABC(4) = 9

A²B² + B²C² + A²C² + 8ABC = 9



(A + B + C)³ = 4³

A³ + B³ + C³ + 6ABC + 3AB² + 3AC² + 3A²B + 3BC² + 3A²C + B²C = 64

22 + 6ABC + 3AB² + 3AC² + 3A²B + 3BC² + 3A²C + 3B²C = 64

6ABC + 3AB² + 3AC² + 3A²B + 3BC² + 3A²C + 3B²C = 42

2ABC + AB² + AC² + A²B + BC² + A²C + B²C = 14

2ABC + AB(A + B) + AC(A + C) + BC(B + C) = 14

2ABC + AB(A + B + C - C) + AC(A + B + C - B) + BC(A + B + C - A) = 14

2ABC + AB(A + B + C) + AC(A + B + C) + BC(A + B + C) - ABC - ABC - ABC = 14

2ABC + (A + B + C)(AB + AC + BC) - 3ABC = 14

(A + B + C)(AB + AC + BC) - ABC = 14

(4)(3) - ABC = 14

12 - ABC = 14

ABC = -2



from above:

A²B² + B²C² + A²C² + 8ABC = 9

A²B² + B²C² + A²C² + 8(-2) = 9

A²B² + B²C² + A²C² - 16 = 9

A²B² + B²C² + A²C² = 25



(A² + B² + C²)(A² + B² + C²) = 10²

A⁴ + B⁴ + C⁴ + 2A²B² + 2B²C² + 2A²C² = 100

A⁴ + B⁴ + C⁴ + 2(A²B² + B²C² + A²C²) = 100

A⁴ + B⁴ + C⁴ + 2(25) = 100

A⁴ + B⁴ + C⁴ + 50 = 100

A⁴ + B⁴ + C⁴ = 50 <<<<<<===== ANSWER

Answer 2

a + b = c is undemanding addition. a^2 + b^2 = c^2 is a geometrical calculation (Pythagorean theorem) so i'm attaching a link b/c the visuals are powerful). a^3+b^3 = c^3 has no geometric relationship. additionally powers better than 2 have been factually shown to not carry this relationship. it relatively is prevalent as Fermat's final Theorem - it relatively is likewise greater complicated and excellent defined in the linked link.

Answer 3

I think you're right. They should be 0, 1, and 3.