By definition, the roots of an equation, substituted in
the equation, cancel out the equation.
We'll move all terms
to one side:
x^2 + 5x - 7 =
0
We'll substitute x1 and x2 in the original
equation:
x1^2 + 5x1 - 7 = 0
(1)
x2^2 + 5x2 - 7 = 0
(2)
We'll add (1) and
(2):
(x1^2 + x2^2) + 5(x1 + x2) - 14 =
0
We'll use Viete's relations to express the sum x1 +
x2:
x1 + x2 = -5
(x1^2
+ x2^2) + 5*(-5) - 14 = 0
(x1^2 + x2^2) - 25 - 14 =
0
We'll combine like
roots:
(x1^2 + x2^2) - 39 =
0
We'll add 39 both
sides:
(x1^2 + x2^2) =
39
Now, we'll re-write the equations (1) and
(2):
x1^2 + 5x1 - 7 = 0
(1)
x2^2 + 5x2 - 7 = 0
(2)
We'll multiply (1) by x1 and (2) by
x2:
x1^3 + 5x1^2 - 7x1 = 0
(3)
x2^3 + 5x2^2 - 7x2 = 0
(4)
We'll add (3) and
(4):
(x1^3 + x2^3) + 5(x1^2 + x2^2) - 7(x1 + x2) =
0
(x1^3 + x2^3) = 7(x1 + x2) - 5(x1^2
+ x2^2)
(x1^3 + x2^3) = 7*(-5) -
5*(39)
(x1^3 + x2^3) = -35 -
195
(x1^3 + x2^3) =
230
The sum of the cubes of the roots of the
given equation is(x1^3 + x2^3) = 230.
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