Question

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Answer 1

Answer:

Answer:

$\frac{a(a^2-4b)(a^2-b)}{b}$

Give than

α and β are zeroes of polynomial x² - ax + b.

Sum of zeroes = -(Coefficient of x/Coefficient of x²)

α + β = -(-a/1)

α + β = a

Product of zeroes = (Constant Term/Coefficient of x²)

αβ = b/1

αβ = b

Now,

(α - β)² = (α + β)² - 4αβ

(α - β)² = a² - 4b

α² + β² = (α + β)² - 2αβ

α² + β² = a² - 2b

$</p><p>\begin{gathered}{\alpha}^2(\frac{{\alpha}^2}{\beta}-\beta)+{\beta}^2(\frac{{\beta}^2}{\alpha}-\alpha)\\\;\\={\alpha}^2(\frac{{\alpha}^2-{\beta}^2}{\beta})+{\beta}^2(\frac{{\beta}^2-{\alpha}^2}{\alpha})\end{gathered}$

$\begin{gathered}={\alpha}^2(\frac{{\alpha}^2-{\beta}^2}{\beta})-{\beta}^2(\frac{{\alpha}^2-{\beta}^2}{\alpha})\\\;\\=\frac{{\alpha}^3({\alpha}^2-{\beta}^2)-{\beta}^3({\alpha}^2-{\beta}^2)}{\alpha\beta}\\\;\\=\frac{({\alpha}^2-{\beta}^2)({\alpha}^3-{\beta}^3)}{\alpha\beta}\end{gathered}$

$\begin{gathered}=\frac{({\alpha}^2-{\beta}^2)(\alpha^2+\beta^2+\alpha\beta)({\alpha}-{\beta})}{\alpha\beta}\\\;\\=\frac{(\alpha-\beta)({\alpha}+{\beta})(\alpha^2+\beta^2+\alpha\beta)({\alpha}-{\beta})}{\alpha\beta}\\\;\\=\frac{(\alpha-\beta)^2({\alpha}+{\beta})(\alpha^2+\beta^2+\alpha\beta)}{\alpha\beta}\end{gathered}$

$\begin{gathered}=\frac{(a^2-4b)a(a^2-2b+b)}{b}\\\;\\=\frac{a(a^2-4b)(a^2-b)}{b}\end{gathered}$

Thus option 3) is correct.

Note:-

(a³ - b³) = (a - b)(a² + b² + ab)

(a² - b²) = (a + b)(a - b)

(a - b)² = (a + b)² - 4ab