Question

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Answer 1

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I WILL DO IT

p(y) = y3-ky-6

y=3

p(3)=27-3k-6=0

= 21 = 3k

[K = 7]

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Answer 2

Answer:

According to question:-

$p(x) = {y}^{3} - ky - 6 = 0$

One of the zero is 3

$p( 3) = { 3}^{3} - 3k - 6 = 0$

$= 27 - 3k - 6 = 0$

$21 - 3k = 0$

$- 3k = - 21$

$k = \frac{ - 21}{ - 3}$

$k = 7$

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