2021/037) Solve the system of equations

$a + b + c = 7\cdots(1)$

$a^2 + b^2 + c^2 = 21\cdots(2)$

$abc= 8 \cdots(3)$

Solution 

We are given 

$a + b + c = 7\cdots(1)$

$a^2 + b^2 + c^2 = 21\cdots(2)$

$abc= 8 \cdots(3)$

From (1) and (2) we get

$(a+b+c)^2 - (a^2+b^2+ c^2) = 7^2 - 21$

Or $2(ab+bc+ca) = 28$

Or $ab+bc+ca = 14\cdots(4)$

a , b, c are roots of equation

$(x-a)(x-b)(x-c) = 0$

Or $x^3-x^2(a+b+c) + x(ab + bc + ca) - abc= 0$

Putting the values from (1), (3) and (4) we get 

$x^3 - 7x^2 + 14x - 8 = 0$

We know from above that the value becomes zero when x = 1 so x =1 is a factor and (dividing by x- 1) we get

$(x-1)(x^2 - 6x + 8) = 0$

Or $(x-1)(x-2)(x-4) = 0$

So roots are 1,2, 4 and so we have

$\{a,b,c\} = \{1,2,4\}$