if a, b,c are the roots of the cubic equation x³-x+1 then what is the value of a³ +b³+ c³ ???
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You dont need to find what the roots actually are. Of course, you could, but this equation in fact has only one real root, and the other two are complex roots; gnarly stuff.
It's a lot easier to take two variations of what it means for a, b and c to be roots of the equation, and work from there.
Firstly, the statement "a is a root of x³-x+1" means that if you replace x with a, you'll get zero.
So a³-a+1 = 0. And of course the same is true for b and c, so b³-b+1 = 0, and c³-c+1 = 0.
Add all these together to get a³+b³+c³-(a+b+c)+3 = 0, or a³+b³+c³ = a+b+c-3
Secondly, because cubic equations only have three roots, we can factorise x³-x+1, and rewrite it as (x-a)(x-b)(x-c).
Multiplying this out fully gives x³ - (a+b+c)x² + (ab+ac+bc)x - abc.
Now comparing the coefficients in this expression with our original equation tells us that abc = -1, ab+ac+bc = -1, and (most importantly) a+b+c = 0.
So now we can use a³+b³+c³ = a+b+c-3, together with a+b+c = 0, to find that a³+b³+c³ = -3
It's a lot easier to take two variations of what it means for a, b and c to be roots of the equation, and work from there.
Firstly, the statement "a is a root of x³-x+1" means that if you replace x with a, you'll get zero.
So a³-a+1 = 0. And of course the same is true for b and c, so b³-b+1 = 0, and c³-c+1 = 0.
Add all these together to get a³+b³+c³-(a+b+c)+3 = 0, or a³+b³+c³ = a+b+c-3
Secondly, because cubic equations only have three roots, we can factorise x³-x+1, and rewrite it as (x-a)(x-b)(x-c).
Multiplying this out fully gives x³ - (a+b+c)x² + (ab+ac+bc)x - abc.
Now comparing the coefficients in this expression with our original equation tells us that abc = -1, ab+ac+bc = -1, and (most importantly) a+b+c = 0.
So now we can use a³+b³+c³ = a+b+c-3, together with a+b+c = 0, to find that a³+b³+c³ = -3
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NO Idea.... but I did think about it