α, β, γ are zeroes of cubic polynomial kx3 – 5x + 9.
If α3 + β3 + γ3 = 27, find the value of k.
Answers
Step-by-step explanation:
The value of k is -1
Step-by-step explanation:
Given that \alpha, \beta, \gammaα,β,γ are zeroes of cubic polynomial kx^3-5x+9kx
3
−5x+9
Also, \alpha^3+\beta^3+\gamma^3=27α
3
+β
3
+γ
3
=27
we have to find the value of k
kx^3+0x^2-5x+9kx
3
+0x
2
−5x+9
\text{Sum of zeroes=}\frac{-b}{a}=\frac{0}{1}=0Sum of zeroes=
a
−b
=
1
0
=0
\text{Product of zeroes=}\frac{-d}{a}=\frac{-9}{k}=\alpha \beta \gammaProduct of zeroes=
a
−d
=
k
−9
=αβγ
As, by formula
x^3+y^3+z^3=3xyzx
3
+y
3
+z
3
=3xyz if x+y+z=0
Since \alpha+\beta+\gamma=0α+β+γ=0
∴ \alpha^3+\beta^3+\gamma^3=3\alpha \beta \gammaα
3
+β
3
+γ
3
=3αβγ
⇒ 27=3\times \frac{-9}{k}27=3×
k
−9
⇒ k=-1k=−1
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