From the sum of $z^{3} + 3 z^{2} + 5 z + 8$ and $4 z^{3} + 2 z^{2} - 7 z - 2$ subtract $2 z^{3} - 3 z^{2} + z - 4$ .

0

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8 z^{2} - 3 z

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3 z^{3} + 8 z^{2} - 3 z + 10

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Solution

The correct option is C $3 z^{3} + 8 z^{2} - 3 z + 10$
The sum of $z^{3} + 3 z^{2} + 5 z + 8$ and $4 z^{3} + 2 z^{2} - 7 z - 2$ is,
$= z^{3} + 3 z^{2} + 5 z + 8 + 4 z^{3} + 2 z^{2} - 7 z - 2$
$= 5 z^{3} + 5 z^{2} - 2 z + 6$
Now,
Subtracting $2 z^{3} - 3 z^{2} + z - 4$ from $= 5 z^{3} + 5 z^{2} - 2 z + 6$ we get,
$5 z^{3} + 5 z^{2} - 2 z + 6$
$2 z^{3} - 3 z^{2} + z - 4$
- + - +
$3 z^{3} + 8 z^{2} - 3 z + 10$

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| z_{1} | = 2, | z_{2} | = 3, | z_{3} | = 4

| 2 z_{1} + 3 z_{2} + 4 z_{3} | = 10

8 z_{2} z_{3} + 27 z_{3} z_{1} + 64 z_{1} z_{2}

If x + y + z = xyz. Then

$\frac{3 x - x^{3}}{1 - 3 x^{2}}$ + $\frac{3 y - y^{3}}{1 - 3 y^{2}}$ + $\frac{3 z - z^{3}}{1 - 3 z^{3}}$ =

x + y + z = x y z

\frac{3 x - x^{3}}{1 - 3 x^{2}} + \frac{3 y - y^{3}}{1 - 3 y^{2}} + \frac{3 z - z^{3}}{1 - 3 z^{2}} = \frac{3 x - x^{3}}{1 - 3 x^{2}} \cdot \frac{3 y - y^{3}}{1 - 3 y^{2}} \cdot \frac{3 z - z^{3}}{1 - 3 z^{2}}

z^{3} + 2 z^{2} + 3 z + 2 = 0

c

[2, 5]

f (z) = 4 z^{3} - 8 z^{2} + 7 z - 2

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