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EXCELENT! 👏👍🙏💎🥇🌹😍
(x-1)^3+(x-2)^3+(x-3)^3+(x-4)^3+(x-5)^3+(x-6)^3+(x-7)^3+(x-8)^3=2024 x=10 x=1.75±0.25Sqrt[615]i=(7±Sqrt[615]i)/4
X =10.
Answer is wrong. Addition of (2a)^3's is 32a^3 at 6.15
Using the formulae for the sums of 8 natural numbers, their squares and their cubes, we get 8[x^3_27/2x^2+153/2x-162]= 2024. So, x^3-27/2x^2+153/2x-415=0. x=10 is the only real solution.
Yes.
2024=2025-1=45^2-1={9×10/2)^2-1=1^3+2^3+...+9^3=1x-1=9,x-8=2(1^3=1 is subtra cted)x=10
9/2, (9±3i(7)^(1/2))/2I apologize, I solved the εξίσωση (x-1)³+(x-2)³ + ... +(x-8)³ = 0 ..I am sorry.
Quick=> as x > INT (Z); for x=9,or10,or11, ..for x=9; ,=> LHS=(8^3+7^3+. .+1^3)=1296; now take x=10; LHS= (9^3+8^3+...2^3)=9^3+1296-1^3 = 2024=RHS; >x=10{Let S= 2024; x-8=a;=>(a+7)^3+(a+6)^3+..a^3=S;a^3+7a^3+3a^2{Σ(1+2+..7)}+3aΣ(1^2+..+7^2)+Σ(1^3+..+7^3); => 8a^3+84a^2+ 420a+784=2024;>8a^3+84a^2+420a=1240;=> 2a^3+21a^2-105a-310=0;=> (a-2)(2a^2+25a+155)=0; =>a-2=0; >a=2;x-8=a; > x=10.}
EXCELENT! 👏👍🙏💎🥇🌹😍
(x-1)^3+(x-2)^3+(x-3)^3+(x-4)^3+(x-5)^3+(x-6)^3+(x-7)^3+(x-8)^3=2024 x=10 x=1.75±0.25Sqrt[615]i=(7±Sqrt[615]i)/4
X =10.
Answer is wrong. Addition of (2a)^3's is 32a^3 at 6.15
Using the formulae for the sums of 8 natural numbers, their squares and their cubes, we get 8[x^3_27/2x^2+153/2x-162]= 2024. So, x^3-27/2x^2+153/2x-415=0. x=10 is the only real solution.
Yes.
2024=2025-1=45^2-1
={9×10/2)^2-1=1^3+2^3+...+9^3=1
x-1=9,x-8=2(1^3=1 is subtra cted)
x=10
9/2, (9±3i(7)^(1/2))/2
I apologize, I solved the εξίσωση
(x-1)³+(x-2)³ + ... +(x-8)³ = 0 ..
I am sorry.
Quick=> as x > INT (Z); for x=9,or10,or11, ..
for x=9; ,=> LHS=
(8^3+7^3+. .+1^3)
=1296; now take x=10; LHS= (9^3+8^3+...2^3)=9^3+1296-1^3 = 2024=RHS; >x=10
{Let S= 2024; x-8=a;=>(a+7)^3+(a+6)^3+..a^3=S;
a^3+7a^3+3a^2{Σ(1+2+..7)}+3aΣ(1^2+..+7^2)+Σ(1^3+..+7^3); => 8a^3+84a^2+ 420a+784=2024;>8a^3+84a^2+
420a=1240;=> 2a^3+21a^2-105a-310=0;=> (a-2)(2a^2+25a+155)=0; =>a-2=0; >a=2;
x-8=a; > x=10.}