Question

if α,β are zeroes of the polynomial p(x)= 2x^3-7x+3, find the value of α^3+β^3​

Answer 1

The question needs a correction:

$P(x) = 2x^2 - 7x + 3$

A cubic has 3 roots, and the problem would be rather complex if we had to calculate $\alpha^3 + \beta^3 + \gamma^3$

Answer:

$\frac{217}{8}$

Step-by-step explanation:

$(\alpha + \beta)^3 = \alpha^3 + \beta^3 + 3\alpha\beta(\alpha + \beta)$

$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$

We know, $\alpha + \beta = -\frac{b}{a} = -\frac{-7}{2} = \frac{7}{2}$

And, $\alpha\beta = \frac{c}{a} = \frac{3}{2}$

Substituting the values, we get:

$\alpha^3 + \beta^3 = (\frac{7}{2})^3 - 3(\frac{3}{2})(\frac{7}{2}) = \frac{217}{8}$