Question

if α and β are the zeroes of 3x^2+6x+9 then find the value of (alpha + beta)^3 - 3alpha beta /(alpha^2+beta^2)(alpha beta)​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

$\large\underline\pink{Given:-}$

• α and β are the zeroes of 3x^2+6x+9

$\large\underline\pink{To find:-}$

• find the value of (alpha + beta)^3 - 3alpha beta /(alpha^2+beta^2)(alpha beta) ....?

$\large\underline\pink{Solutions:-}$

3x² + 6x + 9

• a = 3
• b = 6
• c = 9

we know that:-

α + β = -b/a

α + β = -(-6)/3

α + β = 2

αβ = c/a

αβ = 9/3

αβ = 3

Squaring both side:-

(α + β)² = (-2)²

α² + β² = 4 - 2αβ

α² + β² = 4 - 2(-2)(3)

α² + β² = 4 - 2(-6)

α² + β² = 4 + 12

α² + β² = 16

Now, The value of (α + β)³ - 3αβ/(α² + β²)(αβ)

(-2)³ - 3(3)/(16)(3)

-8 - 9/48

-17/48

Hence, The value of (α + β)³ - 3αβ/(α² + β²)(αβ) is -17/48.