Question

if α,βgamma are the zeroes of the polynomial x^3+5x-2 then find the value of α^3+β^3+gamma^3

Answer 1

Step-by-step explanation:

Given, ${\alpha,\beta\:and\:\gamma}$ are the zeroes of the polynomial ${x}^{3}+{5x}-{2}$.

We know that, for a cubic polynomial

• Sum of zeroes=$\large\frac{-coefficient\:of\:{x}^{2}}{coefficient\:of\:{x}^{3}}$
• Sum of zeroes taken two at a time =$\large\frac{coefficient\:of\:{x}}{coefficient\:of\:{x}^{3}}$
• Product of zeroes=$\large\frac{-constant\: term}{coefficient\:of\:{x}^{3}}$

So,

${\alpha+\beta+\gamma}={\frac{0}{1}}={0}$

${\alpha}{\beta}+{\beta}{\gamma}+{\alpha}{\gamma}={\frac{5}{1}}={5}$

${\alpha}{\beta}{\gamma}={\frac{-(-2)}{1}}={2}$

Also, we know that

${a}^{3}+{b}^{3}+{c}^{3}={(a+b+c)}{({a}^{2}+{b}^{2}+{c}^{2}}-{(ab+bc+ac))}+{3abc}$

So, from the above results,

${\alpha}^{3}+{\beta}^{3}+{\gamma}^{3}={(\alpha+\beta+\gamma)}{({\alpha}^{2}+{\beta}^{2}+{\gamma}^{2}}-{({\alpha}{\beta}+{\beta}{\gamma}+{\alpha}{\gamma}))}+{3}{\alpha}{\beta}{\gamma}$

$\implies{\alpha}^{3}+{\beta}^{3}+{\gamma}^{3}={(0)}{({\alpha}^{2}+{\beta}^{2}+{\gamma}^{2})}+{3(2)}$

$\implies{\alpha}^{3}+{\beta}^{3}+{\gamma}^{3}=\bold{6}$

Hence, the required value is 6.

Answer 2

Answer:

x³ + 5x – 2

a=1, b=0, c=5, d= - 2

Comparing we get

α+ β+γ = -b/a=0

αβ + αγ+β γ=c/a=5

αβγ=-d/a = 2

Since α,β,γ are zeroes of given equation

So α ³ + 5 α – 2 =0 ⇒ α ³ = 2-5 α…….(1)

β³ + 5β – 2=0 ⇒ β ³ = 2-5 β………(2)

γ ³ + 5 γ – 2 =0⇒ γ ³ = 2-5 γ…….(3)

Adding (1) (2) & (3)

α ³ + β ³ + γ ³ = 6-5(α+ β+γ)

α ³ + β ³ + γ ³ = 6-5(0)=6

α ³ + β ³ + γ ³ = 6