Question

Α, β, γ are zeroes of cubic polynomial kx3 – 5x + 9. if α3 + β3 + γ3 = 27, find the value of k.

Answer 1

Answer:

The value of k is -1

Step-by-step explanation:

Given that $\alpha, \beta, \gamma$ are zeroes of cubic polynomial $kx^3-5x+9$

Also, $\alpha^3+\beta^3+\gamma^3=27$

we have to find the value of k

$kx^3+0x^2-5x+9$

$\text{Sum of zeroes=}\frac{-b}{a}=\frac{0}{1}=0$

$\text{Product of zeroes=}\frac{-d}{a}=\frac{-9}{k}=\alpha \beta \gamma$

As, by formula

$x^3+y^3+z^3=3xyz$ if x+y+z=0

Since $\alpha+\beta+\gamma=0$

∴  $\alpha^3+\beta^3+\gamma^3=3\alpha \beta \gamma$

⇒ $27=3\times \frac{-9}{k}$

⇒ $k=-1$

Answer 2

Answer:

k= -1

Step-by-step explanation:

x³+y³+ z³=3 x y z if x+ y+z= 0

since a +b+z=0

therefore ³a +b³+ z ³=3 aby

27= 3×-9/ k

k= -1