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Question:

**If the zeroes of the polynomial x ^{3} – 3x^{2} + x + 1 are a – b, a, a + b, find a and b.**

Answer:

**Solution:** **The given polynomial is ****x ^{3} – 3x^{2} + x + 1. ………. (I)**

**As (a−b), a and (a+b) are the zeros of the given polynomial. **

**Let ****α = (a−b), β = a, and ** ** = ****(a+b). **

**We know,**

**The cubic polynomial with roots α, β, and ** ** is: **

**x ^{3} – (α + β + **

**)x**

^{2}+ (αβ+ β**+**

**α)x – (αβ**

**) ………. (II)**

**On substituting the values of α = (a−b), β = a, and ** ** = (a+b) in (II). We get,**

**x ^{3} – (a - b + a + a + b)x^{2} + [(a−b)a + a(a+b) + (a+b) (a−b)]x – [a(a−b)(a+b)]**

**= **** x ^{3} – 3ax^{2} + [a^{2} – ab + a^{2} + ab + a^{2} – b^{2}]x – a^{3} – ab^{2}**

**= **** x ^{3} – 3ax^{2} + (3a^{2} – b^{2})x – (a^{3} + ab^{2}) …… (III)**

**Now we can Compare the polynomial ****x ^{3} – 3x^{2} + x + 1 **

**with**

**x**

^{3}– 3ax^{2}+ (3a^{2}– b^{2})x – (a^{3}+ ab^{2}),**we obtain,**

**So, 3a = 3 ** ** a = 1, (****3a ^{2} – b^{2}) = 1 and (a^{3} + ab^{2}) = -1**

**Now on substituting the value of a = 1 in (****3a ^{2} – b^{2}) = 1 then we get**

**(****3x****1 ^{2} – b^{2}) = 1**

** 3 - ****b ^{2} = 1**

**3 - 1**

**= b**

^{2} **b ^{2}**

**= 2**

**b =**

**Hence, a = 1 and b = **

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