Question

plzz ans my question ​

Answer 1

It is given that

alpha =a-b

beta =a

Gama =a+b

See the rest of the solution in pic

Answer 2

$f(x) = x^3 - 3x^2 + x + 1$

Given that zeros of polynomial are a, (a + b) and (a - b) which means that zeros are in A.P.

Sum of zeros

$\mathsf{a + (a + b) + (a - b) = \frac{-(-3)}{1}} \\ \\ \Longrightarrow{\mathsf{3a = 3}} \\ \\ \huge{\boxed{\bold{a = 1}}}$

Product of zeros

$\mathsf{a(a + b)(a - b) = - 1} \\ \\ \mathsf{a(a^2 - b^2) = - 1} \\ \\ \mathsf{a^3 - ab^2 = - 1} \\ \\ \mathsf{1 - b^2 = - 1} \\ \\ \huge{\boxed{\boldb = + 2\:and\:-2}}}$

Sum of zeros = - Coefficient of x²/Coefficient of x³

Product of zeros = - Constant Term/Coefficient of x²