if zero of x3 -3×2+x+1 are a-b,a,a+b, then find the value of a2

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if zero of x3 -3×2+x+1 are a-b,a,a+b, then find the value of a2

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Step-by-step explanation:[tex] {x}^{3} – 3 {x}^{2} + x + 1 \\ \alpha = a – b \: , \: \beta =a \: , \: \gamma =a+b \\ sum \: of \: zeroes: \\ \alpha + \beta + \gamma = \frac{-b}{a} \\ (a-b) + a + (a+b)=\frac{-b}{a} \\ 3a = \frac{ – b}{a} \\ {a}^{2} = \frac{ – b}{3} \\ {a}^{2} = \frac{ – ( – 3)}{3} \\ {a}^{2} = 1[/tex]

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Step-by-step explanation:Correct option is

D

a=0,b=−6

x

2

+(a+1)x+b is the quadratic polynomial.

2 and −3 are the zeros of the quadratic polynomial.

Thus, 2+(−3)=

1

−(a+1)

=>

1

(a+1)

=1

=>a+1=1

=>a=0

Also, 2×(−3)=b

=>b=−6