Question

26/x– If a, ß are zeroes of the polynomial p(x) = 3x2 – 2x - 6, then find
alpha³+beta³ ​

Answer 1

Answer:

116/27

Step-by-step explanation:

We have,

a and b are zeros of 3x² - 2x - 6

then,

sum of zeros

a + b = − coff. of x / coff. of x²

a + b = − (-2) / 3

a + b = 2/3

product of zeros

a * b = constant term / coff. of x²

a * b = -6 / 3

a * b = -2

now, a³ + b³ = (a+b)³ −3ab(a+b)

a³ + b³ = (2/3)³ - 3 * -2 * 2/3

a³ + b³ = 8/27 + 4

a³ + b³ = (8 + 108)/27

a³ + b³ = 116/27