three variable are missing a+b+c=0, a²+b²+c²=3, and a⁴+b⁴+c⁴=? And a^5+b^5+c^5=10
If you want integers, then clearly
a,b,c = ±1, since the squares are all positive and add to 3.
But then a+b+c ≠ 1
So, now we need to check real values.
c = -(a+b), so
a^2 + b^2 + (a+b)^2 = 3
2a^2 + 2ab + 2b^2 = 3
a = ±√2, b = ∓1/√2
so, c = ±(√2 - 1/√2)
But now a^5+b^5+c^5 ≠ 10
So now we are left with some complex values. wolframalpha comes up with some, but somehow I don't think that's what you want:
https://www.wolframalpha.com/input/?i=solve+a%2Bb%2Bc%3D0%2C+a%C2%B2%2Bb%C2%B2%2Bc%C2%B2%3D3%2C++a%5E5%2Bb%5E5%2Bc%5E5%3D10
Yes sir ..........
Can you give the identities of
a²+b²+c²=? ,a⁴+b⁴+c⁴=? And a^5+b^5+c^5=?
Thanks Obleck
Master obleck the original question is
a+b+c=0
a³+b³+c³=3
a⁴+b⁴+c⁴=?
a^5+b^5+c^5=10
Sorry for the typo
makes a big difference.
Using my prior steps, I get one solution as
a = -1
b = (1-√5)/2
c = (1+√5)/2
Now surely you can find a^4+b^4+c^4
Give me the identities you used sir
Thanks
I go it no need thanks I understand how to manipulate it
I knew it wouldn't take you very long.
Mont — Please do not use tutors' names in the space for your name. Thanks.
To find the value of a⁴ + b⁴ + c⁴, you can use the following formula:
(a² + b² + c²)² - 2(a·b + b·c + c·a)·(a² + b² + c²)
Let's substitute the given values into this formula to find the result:
Given:
a + b + c = 0 -- Equation 1
a² + b² + c² = 3 -- Equation 2
From Equation 1, we can express one variable in terms of the other two:
a = -(b + c)
Substituting this into Equation 2, we get:
(-b - c)² + b² + c² = 3
b² + c² + 2bc + b² + c² = 3
2(b² + bc + c²) = 3
b² + bc + c² = 3/2 -- Equation 3
To find a⁴ + b⁴ + c⁴, we can use the above formula, so substituting the given equations:
(a² + b² + c²)² - 2(a·b + b·c + c·a)·(a² + b² + c²) =
(3)² - 2(a·b + b·c + c·a)·(3)
We still need to find the value of a·b + b·c + c·a.
From Equation 1, we have:
a + b + c = 0
Rearranging, we get:
a = -(b + c)
Substituting this into a·b + b·c + c·a, we get:
(-(b + c))·b + b·c + c·(-(b + c)) =
- b² - bc - c² + bc - bc - c² =
- b² - 2bc - 2c²
Substituting this value into the formula, we have:
(3)² - 2(- b² - 2bc - 2c²)·(3) =
9 + 2(b² + 2bc + 2c²)·3 =
9 + 6(b² + 2bc + 2c²) =
9 + 6b² + 12bc + 12c²
Now, let's find the value of a⁵ + b⁵ + c⁵ using the given equation:
a⁵ + b⁵ + c⁵ = (a² + b² + c²)(a³ + b³ + c³) - (a·b + b·c + c·a)(a² + b² + c²)
We still need to find the value of a³ + b³ + c³.
To find it, we can use the following formula:
a³ + b³ + c³ = (a + b + c)((a + b + c)² - (a·b + b·c + c·a))
Substituting the given equation a + b + c = 0, we get:
(a + b + c)((a + b + c)² - (a·b + b·c + c·a)) =
0((0)² - (a·b + b·c + c·a)) =
0 - 0 = 0
Substituting this value into a⁵ + b⁵ + c⁵, we have:
(3)(0) - (- b² - 2bc - 2c²)(3) =
0 - (-3b² - 6bc - 6c²) =
3b² + 6bc + 6c²
Therefore, a⁴ + b⁴ + c⁴ = 9 + 6b² + 12bc + 12c², and a⁵ + b⁵ + c⁵ = 3b² + 6bc + 6c².