so beautiful Well I can use Modular Arithmetic a+b+c is congruent to 0 mod 6 now cube both that part a^3 + b^3 + c^3 is congruent to 3(a+b)(b+c)(c+a) mod 6 but if all variables are odd then it is not divisible by 6 So 3(b+c)(c+a)(a+b) is congruent to 0 mod 6 Hence a^3 + b^3 + c ^3 is 0 mod 6
Firstly, a^3 + b^3 + c^3 is congruent to -𝟑(𝐚+𝐛)(𝐛+𝐜)(𝐜+𝐚) 𝐦𝐨𝐝 𝟔. As all variables cannot be odd , one must be even and hence the difference of (a+b+c) and that number must be even, and hence 𝟑(𝐛+𝐜)(𝐜+𝐚)(𝐚+𝐛) 𝐢𝐬 𝐜𝐨𝐧𝐠𝐫𝐮𝐞𝐧𝐭 𝐭𝐨 𝟎 𝐦𝐨𝐝 𝟔, which implies 𝐚^𝟑 + 𝐛^𝟑 + 𝐜 ^𝟑 𝐢𝐬 𝟎 𝐦𝐨𝐝 𝟔
I yhink the same but this is true yhan three odd sum ate not divisible by 6 when i see the question i just reminded of this identity taught in class 9th
without watching the video, here is my method: First of all, note the standard factorisation (note: p3 refers to p cubed p2 is p squared) note that, a3 + b3 + c3 = (a+b+c)(a2 + b2 + c2 - ab - bc - ca) + 3abc Now, it suffices to show that 6 divides 3abc. Now, since 6 divides a + b +c it implies that, 2 must divide a + b + c. We claim: At least one of a, b, c must be even. pf: FTSOC, assume that none of a, b, c are even. Therefore they are odd. Now, their sum is also odd, as a + b + c = 3 (mod 2) = 1 (mod 2). So, this implies 2 does not divide a + b + c. Which violates our given condition. Therefore our initial assumption stands false, and thus, one of a, b, c is even. WLOG, assume a is even. Now, that means there exists k in Z+ such that, 2k = a. Now, 3abc = 6kbc. So, we have 6 divides 3abc, and we are done.
so beautiful
Well I can use Modular Arithmetic
a+b+c is congruent to 0 mod 6
now cube both that part
a^3 + b^3 + c^3 is congruent to 3(a+b)(b+c)(c+a) mod 6
but if all variables are odd then it is not divisible by 6
So 3(b+c)(c+a)(a+b) is congruent to 0 mod 6
Hence a^3 + b^3 + c ^3 is 0 mod 6
Firstly, a^3 + b^3 + c^3 is congruent to -𝟑(𝐚+𝐛)(𝐛+𝐜)(𝐜+𝐚) 𝐦𝐨𝐝 𝟔. As all variables cannot be odd , one must be even and hence the difference of (a+b+c) and that number must be even, and hence 𝟑(𝐛+𝐜)(𝐜+𝐚)(𝐚+𝐛) 𝐢𝐬 𝐜𝐨𝐧𝐠𝐫𝐮𝐞𝐧𝐭 𝐭𝐨 𝟎 𝐦𝐨𝐝 𝟔, which implies 𝐚^𝟑 + 𝐛^𝟑 + 𝐜 ^𝟑 𝐢𝐬 𝟎 𝐦𝐨𝐝 𝟔
Thanks a lot to say that I can also understand maths
You are most welcome
I yhink the same but this is true yhan three odd sum ate not divisible by 6 when i see the question i just reminded of this identity taught in class 9th
Excellent reasoning
Glad you think so!
without watching the video, here is my method:
First of all, note the standard factorisation (note: p3 refers to p cubed p2 is p squared)
note that, a3 + b3 + c3 = (a+b+c)(a2 + b2 + c2 - ab - bc - ca) + 3abc
Now, it suffices to show that 6 divides 3abc.
Now, since 6 divides a + b +c it implies that, 2 must divide a + b + c.
We claim: At least one of a, b, c must be even.
pf: FTSOC, assume that none of a, b, c are even. Therefore they are odd. Now, their sum is also odd, as a + b + c = 3 (mod 2) = 1 (mod 2). So, this implies 2 does not divide a + b + c. Which violates our given condition. Therefore our initial assumption stands false, and thus, one of a, b, c is even.
WLOG, assume a is even. Now, that means there exists k in Z+ such that, 2k = a.
Now, 3abc = 6kbc.
So, we have 6 divides 3abc, and we are done.