if a*a*a+b*b*b+c*c*c=3abc then calculate value of a+b+c?

Answer 1

Answer = ± 1
a + b + c = ± 3

Answer 2

There are two cases I can find: either at least one of the numbers is 0, or none of the numbers is 0.

If a = 0, then
a³ + b³ + c³ = 3abc
0³ + b³ + c³ = 3(0)bc
b³ + c³ = 0
b³ = -c³
b = -c.

If one of the numbers is 0, then the other two are opposites, and
a + b + c = 0.

In the case of non-zero triplets:
Let k = a + b + c
k³ = (a + b + c)³
k³ = a³ + b³ + c³ + 3a²b + 3a²c +3b²c + 3ab² + 3ac² +3bc² + 6abc
k³ = 3abc + 3a²b + 3a²c +3b²c + 3ab² + 3ac² +3bc² + 6abc
k³ = 9abc + 3a²b + 3a²c +3b²c + 3ab² + 3ac² +3bc²
k³ = 3(3abc + a²b + a²c +b²c + ab² + ac² +bc²)
k³ = 3(a²b + a²c +abc +b²c + ab² +abc + ac² +bc² + abc)
k³ = 3[(a²b + a²c +abc) + (b²c + ab² +abc) + (ac² +bc² + abc)]
k³ = 3[a(ab + ac +bc) + b(bc + ab +ac) + c(ac +bc + ab)]
k³ = 3(a + b + c)(ab + ac +bc)
k³ = 3k(ab + ac +bc)
k² = 3(ab + ac + bc)

(a + b + c)² = 3(ab + ac + bc)
a² + b² + c² + 2ab + 2ac + 2bc = 3ab + 3ac + 3bc
a² + b² + c² = ab + ac + bc

a² - (ab + ac) + (b² - bc + c²) = 0
a² - a(b + c) + (b² - bc + c²) = 0

a = {(b + c) ± √[(-b - c)² - 4(1)(b² - bc + c²)]} / 2
a = {(b + c) ± √[b² + 2bc + c² - 4b² + 4bc - 4c²)]} / 2
a = {(b + c) ± √[-3b² + 6bc -3c²]} / 2
a = {(b + c) ± √[-3(b² - 2bc + c²)]} / 2
a = {(b + c) ± √[-3(b - c)²]} / 2
a = {(b + c) ± 3|b - c|i} / 2

For a to be a real number, then |b - c| = 0,
therefore b = c.

a = {(b + c) ± 3|b - c|i} / 2
a = {(b + b) ± 3|b - b|i} / 2
a = {2b} / 2
a = b.

Therefore, a = b = c.

a³ + b³ + c³ = 3abc
a³ + a³ + a³ = 3a³
3a³ = 3a³
0 = 0; "a" can be any real number.

a³ + b³ + c³ = 3abc
has solutions for any real numbers a, b, and c, where a = b = c.
The value of (a + b + c) = 3a.

Answer 3

Use this theoram.

[a1 +a2 + a3 + a4 + ... an ] / n>= [a1*a2*a3*a4...*an ]^(1/n)

for all a1,a2,a3....an >= 0

But here the condition is not given. But its worth giving try... :)

The theoram also says that if the equality holds then all a1,a2,a3... are equal to 1

now lets apply the theoram

[ a^3 + b^3 + c^3 ] /3 >= [ a^3*b^3*c^3 ]^(1/3) ------- (z)

==> a^3 + b^3 + c^3 >= 3abc

but the question says a^3 + b^3 + c^3 = 3abc

which means the equality holds in (z)
no a = b = c

now its clear that a = b = c = 1

But we made an assumptin that a,b,c are positive...
Therefore when a = b = c = 1

a + b + c gives 3 as the answer...

a + b + c = 3

There might be some other answers for sure... :) if a = b = c = -1
also the condition given in the quesiotn is satisfied... :)

therefore another answer is -1 :) :)


some possibloe answrs are 1,-1,0

Answer 4

a+b+c=1/3 of 3abc

Answer 5

a+b+c = 3

Answer 6

the expression can be written as
(asume the number written on the right side as a post script)

(a+b+c)3 = a3+b3+c3+3abc
given:
(a+b+c)3 = 3abc+3abc
therefore

(a+b+c)3 = 6 abc
and
a+b+c =root 6 abc

Answer 7

a+b+c= 1/3abc

Answer 8

a+b+c= 1/3abc

Answer 9

If (a +b+c)= 0
If a+b= -c ......................i
cubing both sides
If (a+b)^3= (-c)^3
If a^3+3ab(a+b) +b^3= -c^3
If a^3 +3ab(-c) +b^3 =-c^3
On simplifying we get
If a^3 -3abc +b^3 +c^3=0
If a^3 +b^3 +c^3=3abc
But it is given, therefore a+b+c=0


Therefore a^3+b^3 +c^3 = 3abc

Answer 10

This equation will hold true anytime a=b=c so, there are infinite solutions.

You didn't say if a, b, and c had to be integers.

Now, if you require a, b, and c to all be different integers, it becomes an interesting problem.

One solution to that would be -1, 0, 1 and you get zero for both sides of the equation and zero for the sum.