# If a^2+b^2+c^2 = 1 and a^3+b^3+c^3 = 1 then what is the value of a+b+c.

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`a^2+b^2+c^2 = 1 ----(1)`

`a^3+b^3+c^3 = 1 ---(2)`

From (1) we can say that;

`-1<=a<=1`

`-1<=b<=1`

`-1<=c<=1`

If` 0<=a<=1` then` a^2>=a^3` (a^2 = a^3 when a = 1 and a = 0)

Assume that a = 0.2

`a^2 = 0.04`

`a^3 = 0.008`

`a^2>a^3`

If `-1<a<=0` then` a^2>=a^3` (a^2 = a^3 when a = 0)

Assume a = -0.02

`a^2 = 0.04`

`a^3 = -0.008`

`a^2>a^3`

So similarly we can say the same for b and c also.

Now we can say;

`a^2+b^2+c^2>=a^3+b^3+c^3`

It is given that `a^2+b^2+c^2 = a^3+b^3+c^3 = 1`

From the above explanation `a^3 = a^2` only if a = 0 and a = 1

This is same for b and c also.

`a^2+b^2+c^2 = 1 = a^3+b^3+c^3`

To get the above we need two of these to be 0 and one to be 1.

For eg. say a = 0,b = 1 and c = 0. Then` a+b+c = 0+1+0 = 1`

*So the answer is a+b+c = 1*

You should use the following special products, such that:

`a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+ac+bc)`

`1 = (a+b+c)^2 - 2(ab+ac+bc)`

`(a+b+c)^3 = (a^2+b^2+c^2+2(ab+ac+bc))(a+b+c) = a^3 + a^2b + a^2c + b^2a + b^3 + b^2c + ac^2 + bc^2 + c^3 + 2(ab+ac+bc)(a+b+c)`

`a^3+b^3+c^3 = (a+b+c)^3 - 2(ab+ac+bc)(a+b+c) - a^2b - a^2c - b^2a - b^2c - ac^2 - bc^2`

`1 = (a+b+c)^3 -3a^2b - 3ab^2 - 3a^2c - 3ac^2 - 3bc^2 - 3b^2c - 6abc`

Since `a^3+b^3+c^3 = a^2+b^2+c^2 = 1` yields:

`(a+b+c)^3 - 2(ab+ac+bc)(a+b+c) - a^2b - a^2c - b^2a - b^2c - ac^2 - bc^2 = (a+b+c)^2 - 2(ab+ac+bc)`

`(a+b+c)^3 - 2(ab+ac+bc)(a+b+c) - (a+b+c)^2 + 2(ab+ac+bc) = a^2b + a^2c + b^2a + b^2c + ac^2 + bc^2`

`(a+b+c)^2(a+b+c-1) - 2(ab+ac+bc)(a+b+c-1) = a^2b + a^2c + b^2a + b^2c + ac^2 + bc^2`

Factoring out `(a+b+c-1)` yields:

`(a+b+c-1)((a+b+c)^2 - 2(ab+ac+bc)) = a^2b + a^2c + b^2a + b^2c + ac^2 + bc^2`

`(a+b+c-1)(a^2+b^2+c^2) = a^2b + a^2c + b^2a + b^2c + ac^2 + bc^2`

Since `a^2+b^2+c^2 = 1` yields:

`a+b+c-1 = a^2b + a^2c + b^2a + b^2c + ac^2 + bc^2`

`a+b+c = 1 + a^2b + a^2c + b^2a + b^2c + ac^2 + bc^2`

`a+b+c =a^2+b^2+c^2 + a^2b + a^2c + b^2a + b^2c + ac^2 + bc^2`

`a+b+c = a^2(1 + b + c) + b^2(1 + a + c) + c^2(1 + a + b)`

**Hence, evaluating `a+b+c` yields **`a+b+c = a^2(1 + b + c) + b^2(1 + a + c) + c^2(1 + a + b).`